limit of an infinite sum $y_n$











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Let $displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}$. Find $displaystylelim_{nrightarrowinfty}y_n$



I see that the terms in $y_n$ are lie between $displaystylefrac{1}{n}$ and $displaystylefrac{1}{n+1}$.
I am unable to form the telescoping series to get the limit.



P.S. Sorry if this is too basic.






real-analysis sequences-and-series







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    up vote
    -1
    down vote

    favorite












    Let $displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}$. Find $displaystylelim_{nrightarrowinfty}y_n$



    I see that the terms in $y_n$ are lie between $displaystylefrac{1}{n}$ and $displaystylefrac{1}{n+1}$.
    I am unable to form the telescoping series to get the limit.



    P.S. Sorry if this is too basic.






    real-analysis sequences-and-series







    share|cite|improve this question


























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      Let $displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}$. Find $displaystylelim_{nrightarrowinfty}y_n$



      I see that the terms in $y_n$ are lie between $displaystylefrac{1}{n}$ and $displaystylefrac{1}{n+1}$.
      I am unable to form the telescoping series to get the limit.



      P.S. Sorry if this is too basic.






      real-analysis sequences-and-series







      share|cite|improve this question















      Let $displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}$. Find $displaystylelim_{nrightarrowinfty}y_n$



      I see that the terms in $y_n$ are lie between $displaystylefrac{1}{n}$ and $displaystylefrac{1}{n+1}$.
      I am unable to form the telescoping series to get the limit.



      P.S. Sorry if this is too basic.






      real-analysis sequences-and-series




      real-analysis sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 23 at 18:59

























      asked Nov 5 at 6:52









      Yadati Kiran

      1,509518




      1,509518






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          Hint:



          You may squeeze $y_n$ as follows:




          • For $k=1, ldots , n$ you have
            $$frac{n^2}{n^3+2n} leq frac{n^2}{n^3+n+k} leq frac{n^2}{n^3+n}$$
            It follows
            $$ncdot frac{n^2}{n^3+2n} leq y_n leq n cdot frac{n^2}{n^3+n}$$


          Now, take the limits on both sides.






          share|cite|improve this answer





















          • Thanks a lot. I got it too !
            – Yadati Kiran
            Nov 5 at 7:24












          • You are welcome. :-)
            – trancelocation
            Nov 5 at 7:26










          • Cauchy's first theorem fir limits can also be used.
            – Devendra Singh Rana
            Nov 13 at 17:16


















          up vote
          1
          down vote













          Just added for your curiosity.



          If you know harmonic numbers,
          $$displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}=n^2 left(H_{n^3+2 n}-H_{n^3+n}right)$$ Now, using, for large $p$
          $$H_p=gamma +log left({p}right)+frac{1}{2 p}-frac{1}{12
          p^2}+Oleft(frac{1}{p^3}right)$$           and continuing with long division or better with Taylor series for large $n$, you should get
          $$y_n=1-frac{3}{2 n^2}+Oleft(frac{1}{n^3}right)$$ For example
          $$y_{10}=frac{6646021071305823047954845}{6748980404790313121758326}approx 0.9847$$ while the above approximation would give $frac{197}{200}=0.9850$.






          share|cite|improve this answer





















            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            Hint:



            You may squeeze $y_n$ as follows:




            • For $k=1, ldots , n$ you have
              $$frac{n^2}{n^3+2n} leq frac{n^2}{n^3+n+k} leq frac{n^2}{n^3+n}$$
              It follows
              $$ncdot frac{n^2}{n^3+2n} leq y_n leq n cdot frac{n^2}{n^3+n}$$


            Now, take the limits on both sides.






            share|cite|improve this answer





















            • Thanks a lot. I got it too !
              – Yadati Kiran
              Nov 5 at 7:24












            • You are welcome. :-)
              – trancelocation
              Nov 5 at 7:26










            • Cauchy's first theorem fir limits can also be used.
              – Devendra Singh Rana
              Nov 13 at 17:16















            up vote
            4
            down vote



            accepted










            Hint:



            You may squeeze $y_n$ as follows:




            • For $k=1, ldots , n$ you have
              $$frac{n^2}{n^3+2n} leq frac{n^2}{n^3+n+k} leq frac{n^2}{n^3+n}$$
              It follows
              $$ncdot frac{n^2}{n^3+2n} leq y_n leq n cdot frac{n^2}{n^3+n}$$


            Now, take the limits on both sides.






            share|cite|improve this answer





















            • Thanks a lot. I got it too !
              – Yadati Kiran
              Nov 5 at 7:24












            • You are welcome. :-)
              – trancelocation
              Nov 5 at 7:26










            • Cauchy's first theorem fir limits can also be used.
              – Devendra Singh Rana
              Nov 13 at 17:16













            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            Hint:



            You may squeeze $y_n$ as follows:




            • For $k=1, ldots , n$ you have
              $$frac{n^2}{n^3+2n} leq frac{n^2}{n^3+n+k} leq frac{n^2}{n^3+n}$$
              It follows
              $$ncdot frac{n^2}{n^3+2n} leq y_n leq n cdot frac{n^2}{n^3+n}$$


            Now, take the limits on both sides.






            share|cite|improve this answer












            Hint:



            You may squeeze $y_n$ as follows:




            • For $k=1, ldots , n$ you have
              $$frac{n^2}{n^3+2n} leq frac{n^2}{n^3+n+k} leq frac{n^2}{n^3+n}$$
              It follows
              $$ncdot frac{n^2}{n^3+2n} leq y_n leq n cdot frac{n^2}{n^3+n}$$


            Now, take the limits on both sides.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 5 at 7:12









            trancelocation

            8,9051521




            8,9051521












            • Thanks a lot. I got it too !
              – Yadati Kiran
              Nov 5 at 7:24












            • You are welcome. :-)
              – trancelocation
              Nov 5 at 7:26










            • Cauchy's first theorem fir limits can also be used.
              – Devendra Singh Rana
              Nov 13 at 17:16


















            • Thanks a lot. I got it too !
              – Yadati Kiran
              Nov 5 at 7:24












            • You are welcome. :-)
              – trancelocation
              Nov 5 at 7:26










            • Cauchy's first theorem fir limits can also be used.
              – Devendra Singh Rana
              Nov 13 at 17:16
















            Thanks a lot. I got it too !
            – Yadati Kiran
            Nov 5 at 7:24






            Thanks a lot. I got it too !
            – Yadati Kiran
            Nov 5 at 7:24














            You are welcome. :-)
            – trancelocation
            Nov 5 at 7:26




            You are welcome. :-)
            – trancelocation
            Nov 5 at 7:26












            Cauchy's first theorem fir limits can also be used.
            – Devendra Singh Rana
            Nov 13 at 17:16




            Cauchy's first theorem fir limits can also be used.
            – Devendra Singh Rana
            Nov 13 at 17:16










            up vote
            1
            down vote













            Just added for your curiosity.



            If you know harmonic numbers,
            $$displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}=n^2 left(H_{n^3+2 n}-H_{n^3+n}right)$$ Now, using, for large $p$
            $$H_p=gamma +log left({p}right)+frac{1}{2 p}-frac{1}{12
            p^2}+Oleft(frac{1}{p^3}right)$$             and continuing with long division or better with Taylor series for large $n$, you should get
            $$y_n=1-frac{3}{2 n^2}+Oleft(frac{1}{n^3}right)$$ For example
            $$y_{10}=frac{6646021071305823047954845}{6748980404790313121758326}approx 0.9847$$ while the above approximation would give $frac{197}{200}=0.9850$.






            share|cite|improve this answer

























              up vote
              1
              down vote













              Just added for your curiosity.



              If you know harmonic numbers,
              $$displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}=n^2 left(H_{n^3+2 n}-H_{n^3+n}right)$$ Now, using, for large $p$
              $$H_p=gamma +log left({p}right)+frac{1}{2 p}-frac{1}{12
              p^2}+Oleft(frac{1}{p^3}right)$$               and continuing with long division or better with Taylor series for large $n$, you should get
              $$y_n=1-frac{3}{2 n^2}+Oleft(frac{1}{n^3}right)$$ For example
              $$y_{10}=frac{6646021071305823047954845}{6748980404790313121758326}approx 0.9847$$ while the above approximation would give $frac{197}{200}=0.9850$.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Just added for your curiosity.



                If you know harmonic numbers,
                $$displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}=n^2 left(H_{n^3+2 n}-H_{n^3+n}right)$$ Now, using, for large $p$
                $$H_p=gamma +log left({p}right)+frac{1}{2 p}-frac{1}{12
                p^2}+Oleft(frac{1}{p^3}right)$$                 and continuing with long division or better with Taylor series for large $n$, you should get
                $$y_n=1-frac{3}{2 n^2}+Oleft(frac{1}{n^3}right)$$ For example
                $$y_{10}=frac{6646021071305823047954845}{6748980404790313121758326}approx 0.9847$$ while the above approximation would give $frac{197}{200}=0.9850$.






                share|cite|improve this answer












                Just added for your curiosity.



                If you know harmonic numbers,
                $$displaystyle y_n=sum_{k=1}^n frac{n^2}{n^3+n+k}=n^2 left(H_{n^3+2 n}-H_{n^3+n}right)$$ Now, using, for large $p$
                $$H_p=gamma +log left({p}right)+frac{1}{2 p}-frac{1}{12
                p^2}+Oleft(frac{1}{p^3}right)$$                 and continuing with long division or better with Taylor series for large $n$, you should get
                $$y_n=1-frac{3}{2 n^2}+Oleft(frac{1}{n^3}right)$$ For example
                $$y_{10}=frac{6646021071305823047954845}{6748980404790313121758326}approx 0.9847$$ while the above approximation would give $frac{197}{200}=0.9850$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 5 at 8:33









                Claude Leibovici

                118k1156131




                118k1156131






























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