Checking the convergence of various series

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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







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  • 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














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







share|cite|improve this question




















  • 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








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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









A.Γ.

22.6k32656




22.6k32656










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














  • 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










1 Answer
1






active

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? $$ |e^{3x}|<1.$$







share|cite|improve this answer























  • 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













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

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? $$ |e^{3x}|<1.$$







share|cite|improve this answer























  • 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














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? $$ |e^{3x}|<1.$$







share|cite|improve this answer























  • 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












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? $$ |e^{3x}|<1.$$







share|cite|improve this answer














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? $$ |e^{3x}|<1.$$








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























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




















  • 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




















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