Show that any null sequence fulfils a specific limit












2












$begingroup$


Show that any null sequence $(a_n)_{ninmathbb R}, a_nneq 0$ fulfils



$$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}tag{1}$$



Was wondering if my approach is valid.



Let's assume that $lim_{ntoinfty}a_n$ exists and that $lim_{ntoinfty}=0$ then we know that $lim_{ntoinfty} (sqrt{1+a_n} -1)$ also exists. So we can write:



$$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{lim_{ntoinfty}sqrt{1+a_n}-1}{lim_{ntoinfty}a_n}=frac{1}{2}tag{2}$$



Now we know that we can rewrite this to



$$lim_{ntoinfty}sqrt{1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{3}$$



We use the fact that the square root is continuous:



$$lim_{ntoinfty}sqrt{1+a_n}-1=sqrt{lim_{ntoinfty}1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{4}$$



We get



$$sqrt{1+0}-1=frac{1}{2}cdot 0tag{5}$$



So the equation does hold for any null sequence $a_n$.



Question: Is there any flaw here?






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    There must be a flaw, because using the same technique you could prove that $lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n} $ is $17$ or $pi$ or whatever you want.
    $endgroup$
    – Martin R
    Jan 24 at 14:47


















2












$begingroup$


Show that any null sequence $(a_n)_{ninmathbb R}, a_nneq 0$ fulfils



$$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}tag{1}$$



Was wondering if my approach is valid.



Let's assume that $lim_{ntoinfty}a_n$ exists and that $lim_{ntoinfty}=0$ then we know that $lim_{ntoinfty} (sqrt{1+a_n} -1)$ also exists. So we can write:



$$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{lim_{ntoinfty}sqrt{1+a_n}-1}{lim_{ntoinfty}a_n}=frac{1}{2}tag{2}$$



Now we know that we can rewrite this to



$$lim_{ntoinfty}sqrt{1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{3}$$



We use the fact that the square root is continuous:



$$lim_{ntoinfty}sqrt{1+a_n}-1=sqrt{lim_{ntoinfty}1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{4}$$



We get



$$sqrt{1+0}-1=frac{1}{2}cdot 0tag{5}$$



So the equation does hold for any null sequence $a_n$.



Question: Is there any flaw here?






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    There must be a flaw, because using the same technique you could prove that $lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n} $ is $17$ or $pi$ or whatever you want.
    $endgroup$
    – Martin R
    Jan 24 at 14:47
















2












2








2





$begingroup$


Show that any null sequence $(a_n)_{ninmathbb R}, a_nneq 0$ fulfils



$$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}tag{1}$$



Was wondering if my approach is valid.



Let's assume that $lim_{ntoinfty}a_n$ exists and that $lim_{ntoinfty}=0$ then we know that $lim_{ntoinfty} (sqrt{1+a_n} -1)$ also exists. So we can write:



$$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{lim_{ntoinfty}sqrt{1+a_n}-1}{lim_{ntoinfty}a_n}=frac{1}{2}tag{2}$$



Now we know that we can rewrite this to



$$lim_{ntoinfty}sqrt{1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{3}$$



We use the fact that the square root is continuous:



$$lim_{ntoinfty}sqrt{1+a_n}-1=sqrt{lim_{ntoinfty}1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{4}$$



We get



$$sqrt{1+0}-1=frac{1}{2}cdot 0tag{5}$$



So the equation does hold for any null sequence $a_n$.



Question: Is there any flaw here?






sequences-and-series convergence







share|cite|improve this question











$endgroup$




Show that any null sequence $(a_n)_{ninmathbb R}, a_nneq 0$ fulfils



$$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}tag{1}$$



Was wondering if my approach is valid.



Let's assume that $lim_{ntoinfty}a_n$ exists and that $lim_{ntoinfty}=0$ then we know that $lim_{ntoinfty} (sqrt{1+a_n} -1)$ also exists. So we can write:



$$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{lim_{ntoinfty}sqrt{1+a_n}-1}{lim_{ntoinfty}a_n}=frac{1}{2}tag{2}$$



Now we know that we can rewrite this to



$$lim_{ntoinfty}sqrt{1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{3}$$



We use the fact that the square root is continuous:



$$lim_{ntoinfty}sqrt{1+a_n}-1=sqrt{lim_{ntoinfty}1+a_n}-1 = frac{1}{2}lim_{ntoinfty}a_ntag{4}$$



We get



$$sqrt{1+0}-1=frac{1}{2}cdot 0tag{5}$$



So the equation does hold for any null sequence $a_n$.



Question: Is there any flaw here?






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 12:39







xotix

















asked Jan 24 at 12:03









xotixxotix

291411




291411












  • $begingroup$
    There must be a flaw, because using the same technique you could prove that $lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n} $ is $17$ or $pi$ or whatever you want.
    $endgroup$
    – Martin R
    Jan 24 at 14:47




















  • $begingroup$
    There must be a flaw, because using the same technique you could prove that $lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n} $ is $17$ or $pi$ or whatever you want.
    $endgroup$
    – Martin R
    Jan 24 at 14:47


















$begingroup$
There must be a flaw, because using the same technique you could prove that $lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n} $ is $17$ or $pi$ or whatever you want.
$endgroup$
– Martin R
Jan 24 at 14:47






$begingroup$
There must be a flaw, because using the same technique you could prove that $lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n} $ is $17$ or $pi$ or whatever you want.
$endgroup$
– Martin R
Jan 24 at 14:47












3 Answers
3






active

oldest

votes


















2












$begingroup$

Regarding your question whether there is a flaw: You need to be careful with mixing the assumptions and what you actually want to prove. You cannot use $$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}$$
in (2) because this is what you actually want to prove! Even if the following steps are equivalences (and you have to be very careful there, in particular with square roots), it is confusing at best and therefore not a very solid proof. Therefore my advice: Try writing proofs in a way that you start with the assumptions and end up with the result instead of mixing the result in somewhere in the middle and conclude with a tautology. This will save you from a lot of trouble and also makes your proofs much easier to read.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah, I know I could prove it way more elegant. While it is clear that I can't use a claim to prove that claim, I'm not sure if that's really the case here. Could you elaborate?
    $endgroup$
    – xotix
    Jan 24 at 12:45










  • $begingroup$
    Ok, to be more specific: In (2) you are using the claim you want to prove. Then you proceed to (3), (4) and so on and then end up with a true statement. For this to be consistent, all your steps need to be equivalences. This is very susceptible to mistakes as we are usually thinking top-down and you are reversing it here. In this particular case, there is a problem from (3) to (2) as you are dividing by $lim_{ntoinfty}a_n$, which is assumed to be $0$.
    $endgroup$
    – Klaus
    Jan 24 at 12:54










  • $begingroup$
    I think you meant multiplied, no? Anyway, I see your point. Thanks a lot.
    $endgroup$
    – xotix
    Jan 25 at 13:39



















1












$begingroup$

You divide by zero, which is not allowed. If you know a bit of calculus you can find the solution by examining the derivative of the root function at the point x=1.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It's not zero, is it? It's a null sequence, which does kind of have an inverse, no? Not sure if I can call it inverse though.
    $endgroup$
    – xotix
    Jan 24 at 12:41










  • $begingroup$
    You are dividing by the limit, which is zero. Diving a real number by a sequence wouldn't make any sense anyway.
    $endgroup$
    – Jagol95
    Jan 24 at 18:33










  • $begingroup$
    do you know basic single variable calculus?
    $endgroup$
    – Jagol95
    Jan 24 at 18:45










  • $begingroup$
    The solution is constructed to address question like that exactly. When I see stuff like this I start thinking what it exactly means and just treating a null sequence as the actual number 0 seems strange. Don't we actually work in an extended real space where $pm infty$ are symbol so we can work with it, e.g. we can now write $1/0=infty$ etc.? But I see that this won't allow me what I did.
    $endgroup$
    – xotix
    Jan 25 at 13:37



















0












$begingroup$

Multiply numerator and denominator by $$sqrt{1+a_n}+1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That is most times called "the conjugate of the numerator (or denominator)" +1
    $endgroup$
    – DonAntonio
    Jan 24 at 12:05












  • $begingroup$
    Ok, that is shorter, the conjugate.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:06










  • $begingroup$
    i know these words only in connection with complex numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:13












  • $begingroup$
    The meaning is pretty similar: change the sign of the second term...that's all.
    $endgroup$
    – DonAntonio
    Jan 24 at 12:42










  • $begingroup$
    How does this answer the “Question: Is there any flaw here?”
    $endgroup$
    – Martin R
    Jan 24 at 14:30











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Regarding your question whether there is a flaw: You need to be careful with mixing the assumptions and what you actually want to prove. You cannot use $$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}$$
in (2) because this is what you actually want to prove! Even if the following steps are equivalences (and you have to be very careful there, in particular with square roots), it is confusing at best and therefore not a very solid proof. Therefore my advice: Try writing proofs in a way that you start with the assumptions and end up with the result instead of mixing the result in somewhere in the middle and conclude with a tautology. This will save you from a lot of trouble and also makes your proofs much easier to read.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah, I know I could prove it way more elegant. While it is clear that I can't use a claim to prove that claim, I'm not sure if that's really the case here. Could you elaborate?
    $endgroup$
    – xotix
    Jan 24 at 12:45










  • $begingroup$
    Ok, to be more specific: In (2) you are using the claim you want to prove. Then you proceed to (3), (4) and so on and then end up with a true statement. For this to be consistent, all your steps need to be equivalences. This is very susceptible to mistakes as we are usually thinking top-down and you are reversing it here. In this particular case, there is a problem from (3) to (2) as you are dividing by $lim_{ntoinfty}a_n$, which is assumed to be $0$.
    $endgroup$
    – Klaus
    Jan 24 at 12:54










  • $begingroup$
    I think you meant multiplied, no? Anyway, I see your point. Thanks a lot.
    $endgroup$
    – xotix
    Jan 25 at 13:39
















2












$begingroup$

Regarding your question whether there is a flaw: You need to be careful with mixing the assumptions and what you actually want to prove. You cannot use $$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}$$
in (2) because this is what you actually want to prove! Even if the following steps are equivalences (and you have to be very careful there, in particular with square roots), it is confusing at best and therefore not a very solid proof. Therefore my advice: Try writing proofs in a way that you start with the assumptions and end up with the result instead of mixing the result in somewhere in the middle and conclude with a tautology. This will save you from a lot of trouble and also makes your proofs much easier to read.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah, I know I could prove it way more elegant. While it is clear that I can't use a claim to prove that claim, I'm not sure if that's really the case here. Could you elaborate?
    $endgroup$
    – xotix
    Jan 24 at 12:45










  • $begingroup$
    Ok, to be more specific: In (2) you are using the claim you want to prove. Then you proceed to (3), (4) and so on and then end up with a true statement. For this to be consistent, all your steps need to be equivalences. This is very susceptible to mistakes as we are usually thinking top-down and you are reversing it here. In this particular case, there is a problem from (3) to (2) as you are dividing by $lim_{ntoinfty}a_n$, which is assumed to be $0$.
    $endgroup$
    – Klaus
    Jan 24 at 12:54










  • $begingroup$
    I think you meant multiplied, no? Anyway, I see your point. Thanks a lot.
    $endgroup$
    – xotix
    Jan 25 at 13:39














2












2








2





$begingroup$

Regarding your question whether there is a flaw: You need to be careful with mixing the assumptions and what you actually want to prove. You cannot use $$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}$$
in (2) because this is what you actually want to prove! Even if the following steps are equivalences (and you have to be very careful there, in particular with square roots), it is confusing at best and therefore not a very solid proof. Therefore my advice: Try writing proofs in a way that you start with the assumptions and end up with the result instead of mixing the result in somewhere in the middle and conclude with a tautology. This will save you from a lot of trouble and also makes your proofs much easier to read.






share|cite|improve this answer









$endgroup$



Regarding your question whether there is a flaw: You need to be careful with mixing the assumptions and what you actually want to prove. You cannot use $$lim_{ntoinfty}frac{sqrt{1+a_n}-1}{a_n}=frac{1}{2}$$
in (2) because this is what you actually want to prove! Even if the following steps are equivalences (and you have to be very careful there, in particular with square roots), it is confusing at best and therefore not a very solid proof. Therefore my advice: Try writing proofs in a way that you start with the assumptions and end up with the result instead of mixing the result in somewhere in the middle and conclude with a tautology. This will save you from a lot of trouble and also makes your proofs much easier to read.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 12:13









KlausKlaus

2,12711




2,12711












  • $begingroup$
    Yeah, I know I could prove it way more elegant. While it is clear that I can't use a claim to prove that claim, I'm not sure if that's really the case here. Could you elaborate?
    $endgroup$
    – xotix
    Jan 24 at 12:45










  • $begingroup$
    Ok, to be more specific: In (2) you are using the claim you want to prove. Then you proceed to (3), (4) and so on and then end up with a true statement. For this to be consistent, all your steps need to be equivalences. This is very susceptible to mistakes as we are usually thinking top-down and you are reversing it here. In this particular case, there is a problem from (3) to (2) as you are dividing by $lim_{ntoinfty}a_n$, which is assumed to be $0$.
    $endgroup$
    – Klaus
    Jan 24 at 12:54










  • $begingroup$
    I think you meant multiplied, no? Anyway, I see your point. Thanks a lot.
    $endgroup$
    – xotix
    Jan 25 at 13:39


















  • $begingroup$
    Yeah, I know I could prove it way more elegant. While it is clear that I can't use a claim to prove that claim, I'm not sure if that's really the case here. Could you elaborate?
    $endgroup$
    – xotix
    Jan 24 at 12:45










  • $begingroup$
    Ok, to be more specific: In (2) you are using the claim you want to prove. Then you proceed to (3), (4) and so on and then end up with a true statement. For this to be consistent, all your steps need to be equivalences. This is very susceptible to mistakes as we are usually thinking top-down and you are reversing it here. In this particular case, there is a problem from (3) to (2) as you are dividing by $lim_{ntoinfty}a_n$, which is assumed to be $0$.
    $endgroup$
    – Klaus
    Jan 24 at 12:54










  • $begingroup$
    I think you meant multiplied, no? Anyway, I see your point. Thanks a lot.
    $endgroup$
    – xotix
    Jan 25 at 13:39
















$begingroup$
Yeah, I know I could prove it way more elegant. While it is clear that I can't use a claim to prove that claim, I'm not sure if that's really the case here. Could you elaborate?
$endgroup$
– xotix
Jan 24 at 12:45




$begingroup$
Yeah, I know I could prove it way more elegant. While it is clear that I can't use a claim to prove that claim, I'm not sure if that's really the case here. Could you elaborate?
$endgroup$
– xotix
Jan 24 at 12:45












$begingroup$
Ok, to be more specific: In (2) you are using the claim you want to prove. Then you proceed to (3), (4) and so on and then end up with a true statement. For this to be consistent, all your steps need to be equivalences. This is very susceptible to mistakes as we are usually thinking top-down and you are reversing it here. In this particular case, there is a problem from (3) to (2) as you are dividing by $lim_{ntoinfty}a_n$, which is assumed to be $0$.
$endgroup$
– Klaus
Jan 24 at 12:54




$begingroup$
Ok, to be more specific: In (2) you are using the claim you want to prove. Then you proceed to (3), (4) and so on and then end up with a true statement. For this to be consistent, all your steps need to be equivalences. This is very susceptible to mistakes as we are usually thinking top-down and you are reversing it here. In this particular case, there is a problem from (3) to (2) as you are dividing by $lim_{ntoinfty}a_n$, which is assumed to be $0$.
$endgroup$
– Klaus
Jan 24 at 12:54












$begingroup$
I think you meant multiplied, no? Anyway, I see your point. Thanks a lot.
$endgroup$
– xotix
Jan 25 at 13:39




$begingroup$
I think you meant multiplied, no? Anyway, I see your point. Thanks a lot.
$endgroup$
– xotix
Jan 25 at 13:39











1












$begingroup$

You divide by zero, which is not allowed. If you know a bit of calculus you can find the solution by examining the derivative of the root function at the point x=1.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It's not zero, is it? It's a null sequence, which does kind of have an inverse, no? Not sure if I can call it inverse though.
    $endgroup$
    – xotix
    Jan 24 at 12:41










  • $begingroup$
    You are dividing by the limit, which is zero. Diving a real number by a sequence wouldn't make any sense anyway.
    $endgroup$
    – Jagol95
    Jan 24 at 18:33










  • $begingroup$
    do you know basic single variable calculus?
    $endgroup$
    – Jagol95
    Jan 24 at 18:45










  • $begingroup$
    The solution is constructed to address question like that exactly. When I see stuff like this I start thinking what it exactly means and just treating a null sequence as the actual number 0 seems strange. Don't we actually work in an extended real space where $pm infty$ are symbol so we can work with it, e.g. we can now write $1/0=infty$ etc.? But I see that this won't allow me what I did.
    $endgroup$
    – xotix
    Jan 25 at 13:37
















1












$begingroup$

You divide by zero, which is not allowed. If you know a bit of calculus you can find the solution by examining the derivative of the root function at the point x=1.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It's not zero, is it? It's a null sequence, which does kind of have an inverse, no? Not sure if I can call it inverse though.
    $endgroup$
    – xotix
    Jan 24 at 12:41










  • $begingroup$
    You are dividing by the limit, which is zero. Diving a real number by a sequence wouldn't make any sense anyway.
    $endgroup$
    – Jagol95
    Jan 24 at 18:33










  • $begingroup$
    do you know basic single variable calculus?
    $endgroup$
    – Jagol95
    Jan 24 at 18:45










  • $begingroup$
    The solution is constructed to address question like that exactly. When I see stuff like this I start thinking what it exactly means and just treating a null sequence as the actual number 0 seems strange. Don't we actually work in an extended real space where $pm infty$ are symbol so we can work with it, e.g. we can now write $1/0=infty$ etc.? But I see that this won't allow me what I did.
    $endgroup$
    – xotix
    Jan 25 at 13:37














1












1








1





$begingroup$

You divide by zero, which is not allowed. If you know a bit of calculus you can find the solution by examining the derivative of the root function at the point x=1.






share|cite|improve this answer









$endgroup$



You divide by zero, which is not allowed. If you know a bit of calculus you can find the solution by examining the derivative of the root function at the point x=1.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 12:12









Jagol95Jagol95

1637




1637












  • $begingroup$
    It's not zero, is it? It's a null sequence, which does kind of have an inverse, no? Not sure if I can call it inverse though.
    $endgroup$
    – xotix
    Jan 24 at 12:41










  • $begingroup$
    You are dividing by the limit, which is zero. Diving a real number by a sequence wouldn't make any sense anyway.
    $endgroup$
    – Jagol95
    Jan 24 at 18:33










  • $begingroup$
    do you know basic single variable calculus?
    $endgroup$
    – Jagol95
    Jan 24 at 18:45










  • $begingroup$
    The solution is constructed to address question like that exactly. When I see stuff like this I start thinking what it exactly means and just treating a null sequence as the actual number 0 seems strange. Don't we actually work in an extended real space where $pm infty$ are symbol so we can work with it, e.g. we can now write $1/0=infty$ etc.? But I see that this won't allow me what I did.
    $endgroup$
    – xotix
    Jan 25 at 13:37


















  • $begingroup$
    It's not zero, is it? It's a null sequence, which does kind of have an inverse, no? Not sure if I can call it inverse though.
    $endgroup$
    – xotix
    Jan 24 at 12:41










  • $begingroup$
    You are dividing by the limit, which is zero. Diving a real number by a sequence wouldn't make any sense anyway.
    $endgroup$
    – Jagol95
    Jan 24 at 18:33










  • $begingroup$
    do you know basic single variable calculus?
    $endgroup$
    – Jagol95
    Jan 24 at 18:45










  • $begingroup$
    The solution is constructed to address question like that exactly. When I see stuff like this I start thinking what it exactly means and just treating a null sequence as the actual number 0 seems strange. Don't we actually work in an extended real space where $pm infty$ are symbol so we can work with it, e.g. we can now write $1/0=infty$ etc.? But I see that this won't allow me what I did.
    $endgroup$
    – xotix
    Jan 25 at 13:37
















$begingroup$
It's not zero, is it? It's a null sequence, which does kind of have an inverse, no? Not sure if I can call it inverse though.
$endgroup$
– xotix
Jan 24 at 12:41




$begingroup$
It's not zero, is it? It's a null sequence, which does kind of have an inverse, no? Not sure if I can call it inverse though.
$endgroup$
– xotix
Jan 24 at 12:41












$begingroup$
You are dividing by the limit, which is zero. Diving a real number by a sequence wouldn't make any sense anyway.
$endgroup$
– Jagol95
Jan 24 at 18:33




$begingroup$
You are dividing by the limit, which is zero. Diving a real number by a sequence wouldn't make any sense anyway.
$endgroup$
– Jagol95
Jan 24 at 18:33












$begingroup$
do you know basic single variable calculus?
$endgroup$
– Jagol95
Jan 24 at 18:45




$begingroup$
do you know basic single variable calculus?
$endgroup$
– Jagol95
Jan 24 at 18:45












$begingroup$
The solution is constructed to address question like that exactly. When I see stuff like this I start thinking what it exactly means and just treating a null sequence as the actual number 0 seems strange. Don't we actually work in an extended real space where $pm infty$ are symbol so we can work with it, e.g. we can now write $1/0=infty$ etc.? But I see that this won't allow me what I did.
$endgroup$
– xotix
Jan 25 at 13:37




$begingroup$
The solution is constructed to address question like that exactly. When I see stuff like this I start thinking what it exactly means and just treating a null sequence as the actual number 0 seems strange. Don't we actually work in an extended real space where $pm infty$ are symbol so we can work with it, e.g. we can now write $1/0=infty$ etc.? But I see that this won't allow me what I did.
$endgroup$
– xotix
Jan 25 at 13:37











0












$begingroup$

Multiply numerator and denominator by $$sqrt{1+a_n}+1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That is most times called "the conjugate of the numerator (or denominator)" +1
    $endgroup$
    – DonAntonio
    Jan 24 at 12:05












  • $begingroup$
    Ok, that is shorter, the conjugate.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:06










  • $begingroup$
    i know these words only in connection with complex numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:13












  • $begingroup$
    The meaning is pretty similar: change the sign of the second term...that's all.
    $endgroup$
    – DonAntonio
    Jan 24 at 12:42










  • $begingroup$
    How does this answer the “Question: Is there any flaw here?”
    $endgroup$
    – Martin R
    Jan 24 at 14:30
















0












$begingroup$

Multiply numerator and denominator by $$sqrt{1+a_n}+1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That is most times called "the conjugate of the numerator (or denominator)" +1
    $endgroup$
    – DonAntonio
    Jan 24 at 12:05












  • $begingroup$
    Ok, that is shorter, the conjugate.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:06










  • $begingroup$
    i know these words only in connection with complex numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:13












  • $begingroup$
    The meaning is pretty similar: change the sign of the second term...that's all.
    $endgroup$
    – DonAntonio
    Jan 24 at 12:42










  • $begingroup$
    How does this answer the “Question: Is there any flaw here?”
    $endgroup$
    – Martin R
    Jan 24 at 14:30














0












0








0





$begingroup$

Multiply numerator and denominator by $$sqrt{1+a_n}+1$$






share|cite|improve this answer









$endgroup$



Multiply numerator and denominator by $$sqrt{1+a_n}+1$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 12:04









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77k42866




77k42866












  • $begingroup$
    That is most times called "the conjugate of the numerator (or denominator)" +1
    $endgroup$
    – DonAntonio
    Jan 24 at 12:05












  • $begingroup$
    Ok, that is shorter, the conjugate.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:06










  • $begingroup$
    i know these words only in connection with complex numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:13












  • $begingroup$
    The meaning is pretty similar: change the sign of the second term...that's all.
    $endgroup$
    – DonAntonio
    Jan 24 at 12:42










  • $begingroup$
    How does this answer the “Question: Is there any flaw here?”
    $endgroup$
    – Martin R
    Jan 24 at 14:30


















  • $begingroup$
    That is most times called "the conjugate of the numerator (or denominator)" +1
    $endgroup$
    – DonAntonio
    Jan 24 at 12:05












  • $begingroup$
    Ok, that is shorter, the conjugate.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:06










  • $begingroup$
    i know these words only in connection with complex numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 24 at 12:13












  • $begingroup$
    The meaning is pretty similar: change the sign of the second term...that's all.
    $endgroup$
    – DonAntonio
    Jan 24 at 12:42










  • $begingroup$
    How does this answer the “Question: Is there any flaw here?”
    $endgroup$
    – Martin R
    Jan 24 at 14:30
















$begingroup$
That is most times called "the conjugate of the numerator (or denominator)" +1
$endgroup$
– DonAntonio
Jan 24 at 12:05






$begingroup$
That is most times called "the conjugate of the numerator (or denominator)" +1
$endgroup$
– DonAntonio
Jan 24 at 12:05














$begingroup$
Ok, that is shorter, the conjugate.
$endgroup$
– Dr. Sonnhard Graubner
Jan 24 at 12:06




$begingroup$
Ok, that is shorter, the conjugate.
$endgroup$
– Dr. Sonnhard Graubner
Jan 24 at 12:06












$begingroup$
i know these words only in connection with complex numbers
$endgroup$
– Dr. Sonnhard Graubner
Jan 24 at 12:13






$begingroup$
i know these words only in connection with complex numbers
$endgroup$
– Dr. Sonnhard Graubner
Jan 24 at 12:13














$begingroup$
The meaning is pretty similar: change the sign of the second term...that's all.
$endgroup$
– DonAntonio
Jan 24 at 12:42




$begingroup$
The meaning is pretty similar: change the sign of the second term...that's all.
$endgroup$
– DonAntonio
Jan 24 at 12:42












$begingroup$
How does this answer the “Question: Is there any flaw here?”
$endgroup$
– Martin R
Jan 24 at 14:30




$begingroup$
How does this answer the “Question: Is there any flaw here?”
$endgroup$
– Martin R
Jan 24 at 14:30


















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