Checking the convergence of various series
1
I was trying to find the convergence of 3 series, I don't know if the procedure I used is correct.
$$sum_{n=1}^infty (-1)^n cos(1/n) $$I had no idea how to proceed with this excercise
$$sum_{n=1}^infty (2^n +4^n)/(2e)^n ∼ sum_{n=1}^infty = 4^n / e^n = (4/e)^n$$ I used comparision test, so $4/e > 1$ so the series diverges- When $xinmathbb{R}$ $$sum_{n=1}^infty log(1+(1/n^frac{x}{2}) ∼sum_{n=1}^infty 1/n^frac{x}{2} $$ The series converges if $x/2 > 1, x>2$
- When $xinmathbb{R}$ $$sum_{n=1}^infty e^{3nx}$$ $sqrt[n]{e^{3nx}} = e^{3x} = (e^3)^x$ I used te root test so the series converges if $e^3 leq t$ for $tin (0,1)$, which is not true so the series diverges.
calculus sequences-and-series convergence
edited 2 days ago
A.Γ.
22.6k32656
asked 2 days ago
El Bryan
397
add a comment |
1
I was trying to find the convergence of 3 series, I don't know if the procedure I used is correct.
$$sum_{n=1}^infty (-1)^n cos(1/n) $$I had no idea how to proceed with this excercise
$$sum_{n=1}^infty (2^n +4^n)/(2e)^n ∼ sum_{n=1}^infty = 4^n / e^n = (4/e)^n$$ I used comparision test, so $4/e > 1$ so the series diverges- When $xinmathbb{R}$ $$sum_{n=1}^infty log(1+(1/n^frac{x}{2}) ∼sum_{n=1}^infty 1/n^frac{x}{2} $$ The series converges if $x/2 > 1, x>2$
- When $xinmathbb{R}$ $$sum_{n=1}^infty e^{3nx}$$ $sqrt[n]{e^{3nx}} = e^{3x} = (e^3)^x$ I used te root test so the series converges if $e^3 leq t$ for $tin (0,1)$, which is not true so the series diverges.
calculus sequences-and-series convergence
edited 2 days ago
A.Γ.
22.6k32656
asked 2 days ago
El Bryan
397
1
You probably shouldn't ask multiple questions in one post.
– B. Goddard
2 days ago
You mean it is better that I ask a question for each exercise?
– El Bryan
2 days ago
Yes. And probably not all at once. Ask a question and learn from it, and maybe the next question won't be necessary.
– B. Goddard
2 days ago
Ok, I'll keep that in mind.
– El Bryan
2 days ago
add a comment |
1
1
1
I was trying to find the convergence of 3 series, I don't know if the procedure I used is correct.
$$sum_{n=1}^infty (-1)^n cos(1/n) $$I had no idea how to proceed with this excercise
$$sum_{n=1}^infty (2^n +4^n)/(2e)^n ∼ sum_{n=1}^infty = 4^n / e^n = (4/e)^n$$ I used comparision test, so $4/e > 1$ so the series diverges- When $xinmathbb{R}$ $$sum_{n=1}^infty log(1+(1/n^frac{x}{2}) ∼sum_{n=1}^infty 1/n^frac{x}{2} $$ The series converges if $x/2 > 1, x>2$
- When $xinmathbb{R}$ $$sum_{n=1}^infty e^{3nx}$$ $sqrt[n]{e^{3nx}} = e^{3x} = (e^3)^x$ I used te root test so the series converges if $e^3 leq t$ for $tin (0,1)$, which is not true so the series diverges.
calculus sequences-and-series convergence
edited 2 days ago
A.Γ.
22.6k32656
asked 2 days ago
El Bryan
397
I was trying to find the convergence of 3 series, I don't know if the procedure I used is correct.
$$sum_{n=1}^infty (-1)^n cos(1/n) $$I had no idea how to proceed with this excercise
$$sum_{n=1}^infty (2^n +4^n)/(2e)^n ∼ sum_{n=1}^infty = 4^n / e^n = (4/e)^n$$ I used comparision test, so $4/e > 1$ so the series diverges- When $xinmathbb{R}$ $$sum_{n=1}^infty log(1+(1/n^frac{x}{2}) ∼sum_{n=1}^infty 1/n^frac{x}{2} $$ The series converges if $x/2 > 1, x>2$
- When $xinmathbb{R}$ $$sum_{n=1}^infty e^{3nx}$$ $sqrt[n]{e^{3nx}} = e^{3x} = (e^3)^x$ I used te root test so the series converges if $e^3 leq t$ for $tin (0,1)$, which is not true so the series diverges.
calculus sequences-and-series convergence
calculus sequences-and-series convergence
edited 2 days ago
A.Γ.
22.6k32656
asked 2 days ago
El Bryan
397
edited 2 days ago
A.Γ.
22.6k32656
asked 2 days ago
El Bryan
397
edited 2 days ago
A.Γ.
22.6k32656
edited 2 days ago
A.Γ.
22.6k32656
edited 2 days ago
A.Γ.
22.6k32656
22.6k32656
asked 2 days ago
El Bryan
397
asked 2 days ago
El Bryan
397
asked 2 days ago
El Bryan
397
397
1
You probably shouldn't ask multiple questions in one post.
– B. Goddard
2 days ago
You mean it is better that I ask a question for each exercise?
– El Bryan
2 days ago
Yes. And probably not all at once. Ask a question and learn from it, and maybe the next question won't be necessary.
– B. Goddard
2 days ago
Ok, I'll keep that in mind.
– El Bryan
2 days ago
add a comment |
1
You probably shouldn't ask multiple questions in one post.
– B. Goddard
2 days ago
You mean it is better that I ask a question for each exercise?
– El Bryan
2 days ago
Yes. And probably not all at once. Ask a question and learn from it, and maybe the next question won't be necessary.
– B. Goddard
2 days ago
Ok, I'll keep that in mind.
– El Bryan
2 days ago
1
1
You probably shouldn't ask multiple questions in one post.
– B. Goddard
2 days ago
You probably shouldn't ask multiple questions in one post.
– B. Goddard
2 days ago
You mean it is better that I ask a question for each exercise?
– El Bryan
2 days ago
You mean it is better that I ask a question for each exercise?
– El Bryan
2 days ago
Yes. And probably not all at once. Ask a question and learn from it, and maybe the next question won't be necessary.
– B. Goddard
2 days ago
Yes. And probably not all at once. Ask a question and learn from it, and maybe the next question won't be necessary.
– B. Goddard
2 days ago
Ok, I'll keep that in mind.
– El Bryan
2 days ago
Ok, I'll keep that in mind.
– El Bryan
2 days ago
add a comment |
1 Answer
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1
Some observations.
What is the following limit? $$lim_{n to infty}left|(-1)^n cos
frac1nright| $$Note that $$ frac{2^n +4^n}{(2e)^n}=frac{1
+2^n}{e^n}=frac{1}{e^n}+left(frac{2}{e}right)^n. $$For which values of $x$ do we have the following inequality? $$ |e^{3x}|<1.$$
answered 2 days ago
Olivier Oloa
108k17176293
1) I guess the limit is equal to +1 or -1 , 2) It was that simple! 3) log 1 < 3x so x>0, is it correct?
– El Bryan
2 days ago
That means that it diverges? Can you please tell me which exercises are wrong?
– El Bryan
2 days ago
First series diverges since its general term does not tend to $0$ as $n to infty$. The second series is convergent, as a sum of two convergents series. The third series, your answer is almost fine: as $n to infty$, $logleft(1+frac1{n^{x/2}}right) sim frac1{n^{x/2}}$ . The last series converges iff $x<0$.
– Olivier Oloa
2 days ago
One last thing, why 1/e^n and (2/e)^n coverges
– El Bryan
2 days ago
If $|q[<1$, then the geometric series $sum q^n$ converges. Note that $1/e^n=(1/e)^n$, $e=2.718cdots$.
– Olivier Oloa
2 days ago
|
show 3 more comments
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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oldest
votes
1
Some observations.
What is the following limit? $$lim_{n to infty}left|(-1)^n cos
frac1nright| $$Note that $$ frac{2^n +4^n}{(2e)^n}=frac{1
+2^n}{e^n}=frac{1}{e^n}+left(frac{2}{e}right)^n. $$For which values of $x$ do we have the following inequality? $$ |e^{3x}|<1.$$
answered 2 days ago
Olivier Oloa
108k17176293
1) I guess the limit is equal to +1 or -1 , 2) It was that simple! 3) log 1 < 3x so x>0, is it correct?
– El Bryan
2 days ago
That means that it diverges? Can you please tell me which exercises are wrong?
– El Bryan
2 days ago
First series diverges since its general term does not tend to $0$ as $n to infty$. The second series is convergent, as a sum of two convergents series. The third series, your answer is almost fine: as $n to infty$, $logleft(1+frac1{n^{x/2}}right) sim frac1{n^{x/2}}$ . The last series converges iff $x<0$.
– Olivier Oloa
2 days ago
One last thing, why 1/e^n and (2/e)^n coverges
– El Bryan
2 days ago
If $|q[<1$, then the geometric series $sum q^n$ converges. Note that $1/e^n=(1/e)^n$, $e=2.718cdots$.
– Olivier Oloa
2 days ago
|
show 3 more comments
1
Some observations.
What is the following limit? $$lim_{n to infty}left|(-1)^n cos
frac1nright| $$Note that $$ frac{2^n +4^n}{(2e)^n}=frac{1
+2^n}{e^n}=frac{1}{e^n}+left(frac{2}{e}right)^n. $$For which values of $x$ do we have the following inequality? $$ |e^{3x}|<1.$$
answered 2 days ago
Olivier Oloa
108k17176293
1) I guess the limit is equal to +1 or -1 , 2) It was that simple! 3) log 1 < 3x so x>0, is it correct?
– El Bryan
2 days ago
That means that it diverges? Can you please tell me which exercises are wrong?
– El Bryan
2 days ago
First series diverges since its general term does not tend to $0$ as $n to infty$. The second series is convergent, as a sum of two convergents series. The third series, your answer is almost fine: as $n to infty$, $logleft(1+frac1{n^{x/2}}right) sim frac1{n^{x/2}}$ . The last series converges iff $x<0$.
– Olivier Oloa
2 days ago
One last thing, why 1/e^n and (2/e)^n coverges
– El Bryan
2 days ago
If $|q[<1$, then the geometric series $sum q^n$ converges. Note that $1/e^n=(1/e)^n$, $e=2.718cdots$.
– Olivier Oloa
2 days ago
|
show 3 more comments
1
1
1
Some observations.
What is the following limit? $$lim_{n to infty}left|(-1)^n cos
frac1nright| $$Note that $$ frac{2^n +4^n}{(2e)^n}=frac{1
+2^n}{e^n}=frac{1}{e^n}+left(frac{2}{e}right)^n. $$For which values of $x$ do we have the following inequality? $$ |e^{3x}|<1.$$
answered 2 days ago
Olivier Oloa
108k17176293
Some observations.
What is the following limit? $$lim_{n to infty}left|(-1)^n cos
frac1nright| $$Note that $$ frac{2^n +4^n}{(2e)^n}=frac{1
+2^n}{e^n}=frac{1}{e^n}+left(frac{2}{e}right)^n. $$For which values of $x$ do we have the following inequality? $$ |e^{3x}|<1.$$
answered 2 days ago
Olivier Oloa
108k17176293
edited 2 days ago
answered 2 days ago
Olivier Oloa
108k17176293
answered 2 days ago
Olivier Oloa
108k17176293
answered 2 days ago
Olivier Oloa
108k17176293
108k17176293
1) I guess the limit is equal to +1 or -1 , 2) It was that simple! 3) log 1 < 3x so x>0, is it correct?
– El Bryan
2 days ago
That means that it diverges? Can you please tell me which exercises are wrong?
– El Bryan
2 days ago
First series diverges since its general term does not tend to $0$ as $n to infty$. The second series is convergent, as a sum of two convergents series. The third series, your answer is almost fine: as $n to infty$, $logleft(1+frac1{n^{x/2}}right) sim frac1{n^{x/2}}$ . The last series converges iff $x<0$.
– Olivier Oloa
2 days ago
One last thing, why 1/e^n and (2/e)^n coverges
– El Bryan
2 days ago
If $|q[<1$, then the geometric series $sum q^n$ converges. Note that $1/e^n=(1/e)^n$, $e=2.718cdots$.
– Olivier Oloa
2 days ago
|
show 3 more comments
1) I guess the limit is equal to +1 or -1 , 2) It was that simple! 3) log 1 < 3x so x>0, is it correct?
– El Bryan
2 days ago
That means that it diverges? Can you please tell me which exercises are wrong?
– El Bryan
2 days ago
First series diverges since its general term does not tend to $0$ as $n to infty$. The second series is convergent, as a sum of two convergents series. The third series, your answer is almost fine: as $n to infty$, $logleft(1+frac1{n^{x/2}}right) sim frac1{n^{x/2}}$ . The last series converges iff $x<0$.
– Olivier Oloa
2 days ago
One last thing, why 1/e^n and (2/e)^n coverges
– El Bryan
2 days ago
If $|q[<1$, then the geometric series $sum q^n$ converges. Note that $1/e^n=(1/e)^n$, $e=2.718cdots$.
– Olivier Oloa
2 days ago
1) I guess the limit is equal to +1 or -1 , 2) It was that simple! 3) log 1 < 3x so x>0, is it correct?
– El Bryan
2 days ago
1) I guess the limit is equal to +1 or -1 , 2) It was that simple! 3) log 1 < 3x so x>0, is it correct?
– El Bryan
2 days ago
That means that it diverges? Can you please tell me which exercises are wrong?
– El Bryan
2 days ago
That means that it diverges? Can you please tell me which exercises are wrong?
– El Bryan
2 days ago
First series diverges since its general term does not tend to $0$ as $n to infty$. The second series is convergent, as a sum of two convergents series. The third series, your answer is almost fine: as $n to infty$, $logleft(1+frac1{n^{x/2}}right) sim frac1{n^{x/2}}$ . The last series converges iff $x<0$.
– Olivier Oloa
2 days ago
First series diverges since its general term does not tend to $0$ as $n to infty$. The second series is convergent, as a sum of two convergents series. The third series, your answer is almost fine: as $n to infty$, $logleft(1+frac1{n^{x/2}}right) sim frac1{n^{x/2}}$ . The last series converges iff $x<0$.
– Olivier Oloa
2 days ago
One last thing, why 1/e^n and (2/e)^n coverges
– El Bryan
2 days ago
One last thing, why 1/e^n and (2/e)^n coverges
– El Bryan
2 days ago
If $|q[<1$, then the geometric series $sum q^n$ converges. Note that $1/e^n=(1/e)^n$, $e=2.718cdots$.
– Olivier Oloa
2 days ago
If $|q[<1$, then the geometric series $sum q^n$ converges. Note that $1/e^n=(1/e)^n$, $e=2.718cdots$.
– Olivier Oloa
2 days ago
|
show 3 more comments
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You probably shouldn't ask multiple questions in one post.
– B. Goddard
2 days ago
You mean it is better that I ask a question for each exercise?
– El Bryan
2 days ago
Yes. And probably not all at once. Ask a question and learn from it, and maybe the next question won't be necessary.
– B. Goddard
2 days ago
Ok, I'll keep that in mind.
– El Bryan
2 days ago