Series convergence from Rudin [duplicate]

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  • Is this correct for Rudin exercise 3.7? Prove the series is convergent

    1 answer



So, in baby Rudin we have the exercise




Prove that the convergence of $sum a_n$ implies the convergence of $$sum fracsqrta_nn$$ if $a_n geq 0$




My Approach



Assume $Sigma a_n$ converges and allow $S_n = sum_i=1^n fracsqrta_ii$.



By Cauchy-Schwarz we have $S_n^2 = (sum_i=1^n fracsqrta_ii)^2 leq sum_i=1^nfrac1i^2 sum_i=1^n a_i $. Thus $lim_ntoinfty S_n^2$ is finite. Consequently, $S_n$ converges.



Is this correct? I'm not sure if that's proper use of Cauchy-Schwarz or if I can apply limits like that. I appreciate the feedback.




real-analysis sequences-and-series proof-verification




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marked as duplicate by amWhy, max_zorn, hardmath, Community♦ Jul 27 at 2:18


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Yes, looks good!
    – A. Pongrácz
    Jul 26 at 22:46










  • duplicate, and also a duplicate at math.stackexchange.com/questions/2624359/…
    – amWhy
    Jul 26 at 22:59














up vote
2
down vote

favorite













This question already has an answer here:



  • Is this correct for Rudin exercise 3.7? Prove the series is convergent

    1 answer



So, in baby Rudin we have the exercise




Prove that the convergence of $sum a_n$ implies the convergence of $$sum fracsqrta_nn$$ if $a_n geq 0$




My Approach



Assume $Sigma a_n$ converges and allow $S_n = sum_i=1^n fracsqrta_ii$.



By Cauchy-Schwarz we have $S_n^2 = (sum_i=1^n fracsqrta_ii)^2 leq sum_i=1^nfrac1i^2 sum_i=1^n a_i $. Thus $lim_ntoinfty S_n^2$ is finite. Consequently, $S_n$ converges.



Is this correct? I'm not sure if that's proper use of Cauchy-Schwarz or if I can apply limits like that. I appreciate the feedback.




real-analysis sequences-and-series proof-verification




share|cite|improve this question













marked as duplicate by amWhy, max_zorn, hardmath, Community♦ Jul 27 at 2:18


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Yes, looks good!
    – A. Pongrácz
    Jul 26 at 22:46










  • duplicate, and also a duplicate at math.stackexchange.com/questions/2624359/…
    – amWhy
    Jul 26 at 22:59












up vote
2
down vote

favorite









up vote
2
down vote

favorite












This question already has an answer here:



  • Is this correct for Rudin exercise 3.7? Prove the series is convergent

    1 answer



So, in baby Rudin we have the exercise




Prove that the convergence of $sum a_n$ implies the convergence of $$sum fracsqrta_nn$$ if $a_n geq 0$




My Approach



Assume $Sigma a_n$ converges and allow $S_n = sum_i=1^n fracsqrta_ii$.



By Cauchy-Schwarz we have $S_n^2 = (sum_i=1^n fracsqrta_ii)^2 leq sum_i=1^nfrac1i^2 sum_i=1^n a_i $. Thus $lim_ntoinfty S_n^2$ is finite. Consequently, $S_n$ converges.



Is this correct? I'm not sure if that's proper use of Cauchy-Schwarz or if I can apply limits like that. I appreciate the feedback.




real-analysis sequences-and-series proof-verification




share|cite|improve this question














This question already has an answer here:



  • Is this correct for Rudin exercise 3.7? Prove the series is convergent

    1 answer



So, in baby Rudin we have the exercise




Prove that the convergence of $sum a_n$ implies the convergence of $$sum fracsqrta_nn$$ if $a_n geq 0$




My Approach



Assume $Sigma a_n$ converges and allow $S_n = sum_i=1^n fracsqrta_ii$.



By Cauchy-Schwarz we have $S_n^2 = (sum_i=1^n fracsqrta_ii)^2 leq sum_i=1^nfrac1i^2 sum_i=1^n a_i $. Thus $lim_ntoinfty S_n^2$ is finite. Consequently, $S_n$ converges.



Is this correct? I'm not sure if that's proper use of Cauchy-Schwarz or if I can apply limits like that. I appreciate the feedback.





This question already has an answer here:



  • Is this correct for Rudin exercise 3.7? Prove the series is convergent

    1 answer





real-analysis sequences-and-series proof-verification





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 22:55
























asked Jul 26 at 22:39









Good Morning Captain

450417




450417




marked as duplicate by amWhy, max_zorn, hardmath, Community♦ Jul 27 at 2:18


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by amWhy, max_zorn, hardmath, Community♦ Jul 27 at 2:18


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • Yes, looks good!
    – A. Pongrácz
    Jul 26 at 22:46










  • duplicate, and also a duplicate at math.stackexchange.com/questions/2624359/…
    – amWhy
    Jul 26 at 22:59
















  • Yes, looks good!
    – A. Pongrácz
    Jul 26 at 22:46










  • duplicate, and also a duplicate at math.stackexchange.com/questions/2624359/…
    – amWhy
    Jul 26 at 22:59















Yes, looks good!
– A. Pongrácz
Jul 26 at 22:46




Yes, looks good!
– A. Pongrácz
Jul 26 at 22:46












duplicate, and also a duplicate at math.stackexchange.com/questions/2624359/…
– amWhy
Jul 26 at 22:59




duplicate, and also a duplicate at math.stackexchange.com/questions/2624359/…
– amWhy
Jul 26 at 22:59










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Your proof is indeed correct.



For the sake of variety, I would suggest an alternative to Cauchy—Schwarz, namely the Am-GM inequality: analogously to what you have written, for any $ngeq 0$,
$$
0leq sum_k=1^n fracsqrta_kk stackreltinyrm (AM-GM)leq
sum_k=1^n fraca_k+frac1k^22 = frac12left( sum_k=1^n a_k + sum_k=1^n frac1k^2right) leq frac12left( sum_k=1^infty a_k + sum_k=1^infty frac1k^2right)
$$
Thus, the partial sums $S_n$ are bounding, and non-decreasing: $(S_n)_n$ converges by the monotone convergence theorem.






share|cite|improve this answer





















  • Ah yes, it's been edited. I appreciate the other approach. Is there a good reference for these commonly used inequalities? I never really encountered AM-GM in class.
    – Good Morning Captain
    Jul 26 at 22:55










  • AM-GM, Cauchy—Schwarz, Minkowski, Jensen, and Holder are quite common; if you feel adventurous, there are plenty others. @GoodMorningCaptain
    – Clement C.
    Jul 26 at 22:58










  • Appreciate the response.
    – Good Morning Captain
    Jul 26 at 22:58










  • @GoodMorningCaptain You're welcome!
    – Clement C.
    Jul 26 at 22:58

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Your proof is indeed correct.



For the sake of variety, I would suggest an alternative to Cauchy—Schwarz, namely the Am-GM inequality: analogously to what you have written, for any $ngeq 0$,
$$
0leq sum_k=1^n fracsqrta_kk stackreltinyrm (AM-GM)leq
sum_k=1^n fraca_k+frac1k^22 = frac12left( sum_k=1^n a_k + sum_k=1^n frac1k^2right) leq frac12left( sum_k=1^infty a_k + sum_k=1^infty frac1k^2right)
$$
Thus, the partial sums $S_n$ are bounding, and non-decreasing: $(S_n)_n$ converges by the monotone convergence theorem.






share|cite|improve this answer





















  • Ah yes, it's been edited. I appreciate the other approach. Is there a good reference for these commonly used inequalities? I never really encountered AM-GM in class.
    – Good Morning Captain
    Jul 26 at 22:55










  • AM-GM, Cauchy—Schwarz, Minkowski, Jensen, and Holder are quite common; if you feel adventurous, there are plenty others. @GoodMorningCaptain
    – Clement C.
    Jul 26 at 22:58










  • Appreciate the response.
    – Good Morning Captain
    Jul 26 at 22:58










  • @GoodMorningCaptain You're welcome!
    – Clement C.
    Jul 26 at 22:58














up vote
2
down vote



accepted










Your proof is indeed correct.



For the sake of variety, I would suggest an alternative to Cauchy—Schwarz, namely the Am-GM inequality: analogously to what you have written, for any $ngeq 0$,
$$
0leq sum_k=1^n fracsqrta_kk stackreltinyrm (AM-GM)leq
sum_k=1^n fraca_k+frac1k^22 = frac12left( sum_k=1^n a_k + sum_k=1^n frac1k^2right) leq frac12left( sum_k=1^infty a_k + sum_k=1^infty frac1k^2right)
$$
Thus, the partial sums $S_n$ are bounding, and non-decreasing: $(S_n)_n$ converges by the monotone convergence theorem.






share|cite|improve this answer





















  • Ah yes, it's been edited. I appreciate the other approach. Is there a good reference for these commonly used inequalities? I never really encountered AM-GM in class.
    – Good Morning Captain
    Jul 26 at 22:55










  • AM-GM, Cauchy—Schwarz, Minkowski, Jensen, and Holder are quite common; if you feel adventurous, there are plenty others. @GoodMorningCaptain
    – Clement C.
    Jul 26 at 22:58










  • Appreciate the response.
    – Good Morning Captain
    Jul 26 at 22:58










  • @GoodMorningCaptain You're welcome!
    – Clement C.
    Jul 26 at 22:58












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Your proof is indeed correct.



For the sake of variety, I would suggest an alternative to Cauchy—Schwarz, namely the Am-GM inequality: analogously to what you have written, for any $ngeq 0$,
$$
0leq sum_k=1^n fracsqrta_kk stackreltinyrm (AM-GM)leq
sum_k=1^n fraca_k+frac1k^22 = frac12left( sum_k=1^n a_k + sum_k=1^n frac1k^2right) leq frac12left( sum_k=1^infty a_k + sum_k=1^infty frac1k^2right)
$$
Thus, the partial sums $S_n$ are bounding, and non-decreasing: $(S_n)_n$ converges by the monotone convergence theorem.






share|cite|improve this answer













Your proof is indeed correct.



For the sake of variety, I would suggest an alternative to Cauchy—Schwarz, namely the Am-GM inequality: analogously to what you have written, for any $ngeq 0$,
$$
0leq sum_k=1^n fracsqrta_kk stackreltinyrm (AM-GM)leq
sum_k=1^n fraca_k+frac1k^22 = frac12left( sum_k=1^n a_k + sum_k=1^n frac1k^2right) leq frac12left( sum_k=1^infty a_k + sum_k=1^infty frac1k^2right)
$$
Thus, the partial sums $S_n$ are bounding, and non-decreasing: $(S_n)_n$ converges by the monotone convergence theorem.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 26 at 22:53









Clement C.

47k33682




47k33682











  • Ah yes, it's been edited. I appreciate the other approach. Is there a good reference for these commonly used inequalities? I never really encountered AM-GM in class.
    – Good Morning Captain
    Jul 26 at 22:55










  • AM-GM, Cauchy—Schwarz, Minkowski, Jensen, and Holder are quite common; if you feel adventurous, there are plenty others. @GoodMorningCaptain
    – Clement C.
    Jul 26 at 22:58










  • Appreciate the response.
    – Good Morning Captain
    Jul 26 at 22:58










  • @GoodMorningCaptain You're welcome!
    – Clement C.
    Jul 26 at 22:58
















  • Ah yes, it's been edited. I appreciate the other approach. Is there a good reference for these commonly used inequalities? I never really encountered AM-GM in class.
    – Good Morning Captain
    Jul 26 at 22:55










  • AM-GM, Cauchy—Schwarz, Minkowski, Jensen, and Holder are quite common; if you feel adventurous, there are plenty others. @GoodMorningCaptain
    – Clement C.
    Jul 26 at 22:58










  • Appreciate the response.
    – Good Morning Captain
    Jul 26 at 22:58










  • @GoodMorningCaptain You're welcome!
    – Clement C.
    Jul 26 at 22:58















Ah yes, it's been edited. I appreciate the other approach. Is there a good reference for these commonly used inequalities? I never really encountered AM-GM in class.
– Good Morning Captain
Jul 26 at 22:55




Ah yes, it's been edited. I appreciate the other approach. Is there a good reference for these commonly used inequalities? I never really encountered AM-GM in class.
– Good Morning Captain
Jul 26 at 22:55












AM-GM, Cauchy—Schwarz, Minkowski, Jensen, and Holder are quite common; if you feel adventurous, there are plenty others. @GoodMorningCaptain
– Clement C.
Jul 26 at 22:58




AM-GM, Cauchy—Schwarz, Minkowski, Jensen, and Holder are quite common; if you feel adventurous, there are plenty others. @GoodMorningCaptain
– Clement C.
Jul 26 at 22:58












Appreciate the response.
– Good Morning Captain
Jul 26 at 22:58




Appreciate the response.
– Good Morning Captain
Jul 26 at 22:58












@GoodMorningCaptain You're welcome!
– Clement C.
Jul 26 at 22:58




@GoodMorningCaptain You're welcome!
– Clement C.
Jul 26 at 22:58


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