Let ${a_n}$ be a sequence of positive numbers and $b_{n} = frac{a_{n}}{(a_{1}+…+a_{n})^{2}}$. Prove...












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Let ${a_n}_{n=1}^{infty}$ be a sequence of positive numbers and let $b_{n} = frac{a_{n}}{(a_{1}+...+a_{n})^{2}}$ for n $inmathbb{N}$. Prove that $sum_{n=1}^{infty}b_{n}$ is a convergent series.
I'm stuck on how to start this problem. I've considered the limit comparison test, but it hasn't worked out for me. I know I can assume ${a_{n}}$ and ${b_{n}}$ are positive, so maybe I need to show ${b_{n}}$ has an upper bound and apply the positive series test. Any help is appreciated. Thank you!






real-analysis calculus







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    4












    $begingroup$


    Let ${a_n}_{n=1}^{infty}$ be a sequence of positive numbers and let $b_{n} = frac{a_{n}}{(a_{1}+...+a_{n})^{2}}$ for n $inmathbb{N}$. Prove that $sum_{n=1}^{infty}b_{n}$ is a convergent series.
    I'm stuck on how to start this problem. I've considered the limit comparison test, but it hasn't worked out for me. I know I can assume ${a_{n}}$ and ${b_{n}}$ are positive, so maybe I need to show ${b_{n}}$ has an upper bound and apply the positive series test. Any help is appreciated. Thank you!






    real-analysis calculus







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      1



      $begingroup$


      Let ${a_n}_{n=1}^{infty}$ be a sequence of positive numbers and let $b_{n} = frac{a_{n}}{(a_{1}+...+a_{n})^{2}}$ for n $inmathbb{N}$. Prove that $sum_{n=1}^{infty}b_{n}$ is a convergent series.
      I'm stuck on how to start this problem. I've considered the limit comparison test, but it hasn't worked out for me. I know I can assume ${a_{n}}$ and ${b_{n}}$ are positive, so maybe I need to show ${b_{n}}$ has an upper bound and apply the positive series test. Any help is appreciated. Thank you!






      real-analysis calculus







      share|cite|improve this question









      $endgroup$




      Let ${a_n}_{n=1}^{infty}$ be a sequence of positive numbers and let $b_{n} = frac{a_{n}}{(a_{1}+...+a_{n})^{2}}$ for n $inmathbb{N}$. Prove that $sum_{n=1}^{infty}b_{n}$ is a convergent series.
      I'm stuck on how to start this problem. I've considered the limit comparison test, but it hasn't worked out for me. I know I can assume ${a_{n}}$ and ${b_{n}}$ are positive, so maybe I need to show ${b_{n}}$ has an upper bound and apply the positive series test. Any help is appreciated. Thank you!






      real-analysis calculus




      real-analysis calculus






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      asked Jan 2 at 3:17









      dorkichardorkichar

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          $begingroup$

          Hint:



          With $S_n = a_1+ ldots + a_n$,



          $$frac{a_n}{S_n^2} leqslant frac{S_n - S_{n-1}}{S_nS_{n-1}} = frac{1}{S_{n-1}}- frac{1}{S_n}$$



          Note that $S_n$ converges either to a finite limit or $+infty$ and in all cases $1/S_n $ converges to a finite limit.






          share|cite|improve this answer











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          • $begingroup$
            Clever approach and definitely makes sense to me. Thank you!
            $endgroup$
            – dorkichar
            Jan 3 at 21:24












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          1 Answer
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          1 Answer
          1






          active

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          active

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          active

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          7












          $begingroup$

          Hint:



          With $S_n = a_1+ ldots + a_n$,



          $$frac{a_n}{S_n^2} leqslant frac{S_n - S_{n-1}}{S_nS_{n-1}} = frac{1}{S_{n-1}}- frac{1}{S_n}$$



          Note that $S_n$ converges either to a finite limit or $+infty$ and in all cases $1/S_n $ converges to a finite limit.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Clever approach and definitely makes sense to me. Thank you!
            $endgroup$
            – dorkichar
            Jan 3 at 21:24
















          7












          $begingroup$

          Hint:



          With $S_n = a_1+ ldots + a_n$,



          $$frac{a_n}{S_n^2} leqslant frac{S_n - S_{n-1}}{S_nS_{n-1}} = frac{1}{S_{n-1}}- frac{1}{S_n}$$



          Note that $S_n$ converges either to a finite limit or $+infty$ and in all cases $1/S_n $ converges to a finite limit.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Clever approach and definitely makes sense to me. Thank you!
            $endgroup$
            – dorkichar
            Jan 3 at 21:24














          7












          7








          7





          $begingroup$

          Hint:



          With $S_n = a_1+ ldots + a_n$,



          $$frac{a_n}{S_n^2} leqslant frac{S_n - S_{n-1}}{S_nS_{n-1}} = frac{1}{S_{n-1}}- frac{1}{S_n}$$



          Note that $S_n$ converges either to a finite limit or $+infty$ and in all cases $1/S_n $ converges to a finite limit.






          share|cite|improve this answer











          $endgroup$



          Hint:



          With $S_n = a_1+ ldots + a_n$,



          $$frac{a_n}{S_n^2} leqslant frac{S_n - S_{n-1}}{S_nS_{n-1}} = frac{1}{S_{n-1}}- frac{1}{S_n}$$



          Note that $S_n$ converges either to a finite limit or $+infty$ and in all cases $1/S_n $ converges to a finite limit.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 3:51

























          answered Jan 2 at 3:44









          RRLRRL

          53.1k52574




          53.1k52574












          • $begingroup$
            Clever approach and definitely makes sense to me. Thank you!
            $endgroup$
            – dorkichar
            Jan 3 at 21:24


















          • $begingroup$
            Clever approach and definitely makes sense to me. Thank you!
            $endgroup$
            – dorkichar
            Jan 3 at 21:24
















          $begingroup$
          Clever approach and definitely makes sense to me. Thank you!
          $endgroup$
          – dorkichar
          Jan 3 at 21:24




          $begingroup$
          Clever approach and definitely makes sense to me. Thank you!
          $endgroup$
          – dorkichar
          Jan 3 at 21:24


















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