Evaluating Summation of Infinite Series












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$begingroup$


Let $a_{1},a_{2},...,a_{n}$ be a monotone increasing sequence of numbers satisfying $sin(a_{k}) = frac{k}{n}$ and $a_{k} leq frac{pi}{2}$ for all $1leq k leq n$. I would like to calculate $limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}$.



This is my attempt so far. First, fix $ninmathbb{N}$. Then, we have the upper bound as follows :
$$frac{1}{n}sum_{k=1}^{n}a_{k} leq frac{pi}{2}$$
Therefore, $limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}leq frac{pi}{2}$



On the other hand, I also have the following inequality:
begin{align*}
frac{1}{n}sum_{k=1}^{n}sin(a_{k})leq frac{1}{n}sum_{k=1}^{n}a_{k}
end{align*}
Moreover, we know that $sin(a_{k}) =frac{k}{n}$ and therefore $frac{1}{n}sum_{k=1}^{n}sin(a_{k})=frac{n+1}{2n}$
Hence, we have $frac{n+1}{2n}leq frac{1}{n}sum_{k=1}^{n}a_{k}$ which implies $$ frac{1}{2} leq limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}$$



Therefore, I obtain $frac{1}{2} leq limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k} leq frac{pi}{2}$.



Now, I am confused how to find a better bound of this sum to apply the squeeze theorem. Any help or hint will be much appreciated! Thank you!






calculus sequences-and-series limits summation-method







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    1












    $begingroup$


    Let $a_{1},a_{2},...,a_{n}$ be a monotone increasing sequence of numbers satisfying $sin(a_{k}) = frac{k}{n}$ and $a_{k} leq frac{pi}{2}$ for all $1leq k leq n$. I would like to calculate $limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}$.



    This is my attempt so far. First, fix $ninmathbb{N}$. Then, we have the upper bound as follows :
    $$frac{1}{n}sum_{k=1}^{n}a_{k} leq frac{pi}{2}$$
    Therefore, $limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}leq frac{pi}{2}$



    On the other hand, I also have the following inequality:
    begin{align*}
    frac{1}{n}sum_{k=1}^{n}sin(a_{k})leq frac{1}{n}sum_{k=1}^{n}a_{k}
    end{align*}
    Moreover, we know that $sin(a_{k}) =frac{k}{n}$ and therefore $frac{1}{n}sum_{k=1}^{n}sin(a_{k})=frac{n+1}{2n}$
    Hence, we have $frac{n+1}{2n}leq frac{1}{n}sum_{k=1}^{n}a_{k}$ which implies $$ frac{1}{2} leq limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}$$



    Therefore, I obtain $frac{1}{2} leq limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k} leq frac{pi}{2}$.



    Now, I am confused how to find a better bound of this sum to apply the squeeze theorem. Any help or hint will be much appreciated! Thank you!






    calculus sequences-and-series limits summation-method







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $a_{1},a_{2},...,a_{n}$ be a monotone increasing sequence of numbers satisfying $sin(a_{k}) = frac{k}{n}$ and $a_{k} leq frac{pi}{2}$ for all $1leq k leq n$. I would like to calculate $limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}$.



      This is my attempt so far. First, fix $ninmathbb{N}$. Then, we have the upper bound as follows :
      $$frac{1}{n}sum_{k=1}^{n}a_{k} leq frac{pi}{2}$$
      Therefore, $limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}leq frac{pi}{2}$



      On the other hand, I also have the following inequality:
      begin{align*}
      frac{1}{n}sum_{k=1}^{n}sin(a_{k})leq frac{1}{n}sum_{k=1}^{n}a_{k}
      end{align*}
      Moreover, we know that $sin(a_{k}) =frac{k}{n}$ and therefore $frac{1}{n}sum_{k=1}^{n}sin(a_{k})=frac{n+1}{2n}$
      Hence, we have $frac{n+1}{2n}leq frac{1}{n}sum_{k=1}^{n}a_{k}$ which implies $$ frac{1}{2} leq limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}$$



      Therefore, I obtain $frac{1}{2} leq limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k} leq frac{pi}{2}$.



      Now, I am confused how to find a better bound of this sum to apply the squeeze theorem. Any help or hint will be much appreciated! Thank you!






      calculus sequences-and-series limits summation-method







      share|cite|improve this question









      $endgroup$




      Let $a_{1},a_{2},...,a_{n}$ be a monotone increasing sequence of numbers satisfying $sin(a_{k}) = frac{k}{n}$ and $a_{k} leq frac{pi}{2}$ for all $1leq k leq n$. I would like to calculate $limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}$.



      This is my attempt so far. First, fix $ninmathbb{N}$. Then, we have the upper bound as follows :
      $$frac{1}{n}sum_{k=1}^{n}a_{k} leq frac{pi}{2}$$
      Therefore, $limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}leq frac{pi}{2}$



      On the other hand, I also have the following inequality:
      begin{align*}
      frac{1}{n}sum_{k=1}^{n}sin(a_{k})leq frac{1}{n}sum_{k=1}^{n}a_{k}
      end{align*}
      Moreover, we know that $sin(a_{k}) =frac{k}{n}$ and therefore $frac{1}{n}sum_{k=1}^{n}sin(a_{k})=frac{n+1}{2n}$
      Hence, we have $frac{n+1}{2n}leq frac{1}{n}sum_{k=1}^{n}a_{k}$ which implies $$ frac{1}{2} leq limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k}$$



      Therefore, I obtain $frac{1}{2} leq limlimits_{ntoinfty}frac{1}{n}sum_{k=1}^{n}a_{k} leq frac{pi}{2}$.



      Now, I am confused how to find a better bound of this sum to apply the squeeze theorem. Any help or hint will be much appreciated! Thank you!






      calculus sequences-and-series limits summation-method




      calculus sequences-and-series limits summation-method






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      asked Jan 11 at 6:42









      Evan William ChandraEvan William Chandra

      628313




      628313






















          1 Answer
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          $begingroup$

          Hint. Let $y=sin(x)$, then your sum is a Riemann sum with respect to the interval $[0,1]$ along the $y$-axis:
          $$frac{1}{n}sum_{k=1}^{n}a_{k}=frac{1}{n}sum_{k=1}^{n}arcsin(k/n).$$
          So what is the limit you are looking for?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I feel so stupid now! How could I not realise this earlier? Thank you so much! Now this becomes easy thanks to your hint!
            $endgroup$
            – Evan William Chandra
            Jan 11 at 6:58











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          1 Answer
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          active

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          active

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          active

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          $begingroup$

          Hint. Let $y=sin(x)$, then your sum is a Riemann sum with respect to the interval $[0,1]$ along the $y$-axis:
          $$frac{1}{n}sum_{k=1}^{n}a_{k}=frac{1}{n}sum_{k=1}^{n}arcsin(k/n).$$
          So what is the limit you are looking for?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I feel so stupid now! How could I not realise this earlier? Thank you so much! Now this becomes easy thanks to your hint!
            $endgroup$
            – Evan William Chandra
            Jan 11 at 6:58
















          2












          $begingroup$

          Hint. Let $y=sin(x)$, then your sum is a Riemann sum with respect to the interval $[0,1]$ along the $y$-axis:
          $$frac{1}{n}sum_{k=1}^{n}a_{k}=frac{1}{n}sum_{k=1}^{n}arcsin(k/n).$$
          So what is the limit you are looking for?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I feel so stupid now! How could I not realise this earlier? Thank you so much! Now this becomes easy thanks to your hint!
            $endgroup$
            – Evan William Chandra
            Jan 11 at 6:58














          2












          2








          2





          $begingroup$

          Hint. Let $y=sin(x)$, then your sum is a Riemann sum with respect to the interval $[0,1]$ along the $y$-axis:
          $$frac{1}{n}sum_{k=1}^{n}a_{k}=frac{1}{n}sum_{k=1}^{n}arcsin(k/n).$$
          So what is the limit you are looking for?






          share|cite|improve this answer











          $endgroup$



          Hint. Let $y=sin(x)$, then your sum is a Riemann sum with respect to the interval $[0,1]$ along the $y$-axis:
          $$frac{1}{n}sum_{k=1}^{n}a_{k}=frac{1}{n}sum_{k=1}^{n}arcsin(k/n).$$
          So what is the limit you are looking for?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 6:52

























          answered Jan 11 at 6:46









          Robert ZRobert Z

          99.8k1068140




          99.8k1068140












          • $begingroup$
            I feel so stupid now! How could I not realise this earlier? Thank you so much! Now this becomes easy thanks to your hint!
            $endgroup$
            – Evan William Chandra
            Jan 11 at 6:58


















          • $begingroup$
            I feel so stupid now! How could I not realise this earlier? Thank you so much! Now this becomes easy thanks to your hint!
            $endgroup$
            – Evan William Chandra
            Jan 11 at 6:58
















          $begingroup$
          I feel so stupid now! How could I not realise this earlier? Thank you so much! Now this becomes easy thanks to your hint!
          $endgroup$
          – Evan William Chandra
          Jan 11 at 6:58




          $begingroup$
          I feel so stupid now! How could I not realise this earlier? Thank you so much! Now this becomes easy thanks to your hint!
          $endgroup$
          – Evan William Chandra
          Jan 11 at 6:58


















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