What value does $sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}$ converge to?












-1












$begingroup$


What value does



$$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}$$
converge to?



Ok so I've tried changing the sum to:



$$sum_{n=1}^{infty} dfrac{1}{6(2n+1)}-dfrac{1}{6(2n+7)}$$



and then writting some values:
$$frac16·(frac13+frac15+frac17dots+frac1{2N+1})-frac16·(frac19+frac1{11}+frac1{13}dots+frac1{2N+7})$$



but I don't know what else I can do to finish it! Any hint or solution?






sequences-and-series convergence







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$endgroup$








  • 1




    $begingroup$
    In those sums you’ve written out, everything cancels except finitely many terms. For example, in the first sum, the very next term inside the $cdots$ is $frac 1 9$ which cancels with the $frac 19$ in the second sum.
    $endgroup$
    – User8128
    Dec 8 '18 at 15:36
















-1












$begingroup$


What value does



$$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}$$
converge to?



Ok so I've tried changing the sum to:



$$sum_{n=1}^{infty} dfrac{1}{6(2n+1)}-dfrac{1}{6(2n+7)}$$



and then writting some values:
$$frac16·(frac13+frac15+frac17dots+frac1{2N+1})-frac16·(frac19+frac1{11}+frac1{13}dots+frac1{2N+7})$$



but I don't know what else I can do to finish it! Any hint or solution?






sequences-and-series convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    In those sums you’ve written out, everything cancels except finitely many terms. For example, in the first sum, the very next term inside the $cdots$ is $frac 1 9$ which cancels with the $frac 19$ in the second sum.
    $endgroup$
    – User8128
    Dec 8 '18 at 15:36














-1












-1








-1


1



$begingroup$


What value does



$$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}$$
converge to?



Ok so I've tried changing the sum to:



$$sum_{n=1}^{infty} dfrac{1}{6(2n+1)}-dfrac{1}{6(2n+7)}$$



and then writting some values:
$$frac16·(frac13+frac15+frac17dots+frac1{2N+1})-frac16·(frac19+frac1{11}+frac1{13}dots+frac1{2N+7})$$



but I don't know what else I can do to finish it! Any hint or solution?






sequences-and-series convergence







share|cite|improve this question









$endgroup$




What value does



$$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}$$
converge to?



Ok so I've tried changing the sum to:



$$sum_{n=1}^{infty} dfrac{1}{6(2n+1)}-dfrac{1}{6(2n+7)}$$



and then writting some values:
$$frac16·(frac13+frac15+frac17dots+frac1{2N+1})-frac16·(frac19+frac1{11}+frac1{13}dots+frac1{2N+7})$$



but I don't know what else I can do to finish it! Any hint or solution?






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 15:33









iggykimiiggykimi

19410




19410








  • 1




    $begingroup$
    In those sums you’ve written out, everything cancels except finitely many terms. For example, in the first sum, the very next term inside the $cdots$ is $frac 1 9$ which cancels with the $frac 19$ in the second sum.
    $endgroup$
    – User8128
    Dec 8 '18 at 15:36














  • 1




    $begingroup$
    In those sums you’ve written out, everything cancels except finitely many terms. For example, in the first sum, the very next term inside the $cdots$ is $frac 1 9$ which cancels with the $frac 19$ in the second sum.
    $endgroup$
    – User8128
    Dec 8 '18 at 15:36








1




1




$begingroup$
In those sums you’ve written out, everything cancels except finitely many terms. For example, in the first sum, the very next term inside the $cdots$ is $frac 1 9$ which cancels with the $frac 19$ in the second sum.
$endgroup$
– User8128
Dec 8 '18 at 15:36




$begingroup$
In those sums you’ve written out, everything cancels except finitely many terms. For example, in the first sum, the very next term inside the $cdots$ is $frac 1 9$ which cancels with the $frac 19$ in the second sum.
$endgroup$
– User8128
Dec 8 '18 at 15:36










3 Answers
3






active

oldest

votes


















7












$begingroup$

Hint: Let's look at the $100$th partial sum. It's good to get some concreteness.



$$frac{1}{6}left(frac{1}{3}+frac{1}{5}+cdots+frac{1}{201}right)-frac{1}{6}left(frac{1}{9}+cdots+frac{1}{205}+frac{1}{207}right).$$



We have a bunch of terms that are repeated: $frac{1}{9}+cdots+frac{1}{201}$ exists in each bracketed portion, so we can simply cancel all of them out to get



$$frac{1}{6}left(frac{1}{3}+frac{1}{5}+frac{1}{7}-frac{1}{203}-frac{1}{205}-frac{1}{207}right).$$



Can you see how to use this line of reasoning to get the answer?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (+1) Nice hint!
    $endgroup$
    – José Carlos Santos
    Dec 8 '18 at 15:40



















3












$begingroup$

Check that
$$left(frac13+frac15+frac17+frac19+frac1{11}+cdots+frac1{2N+1}right)-$$
$$-left(frac19+frac1{11}+cdots+frac1{2N+1}+frac1{2N+3}+frac1{2N+5}+frac1{2N+7}right)=$$
$$=frac13+frac15+frac17-frac1{2N+3}-frac1{2N+5}-frac1{2N+7}.$$



And if you take the limit as $Ntoinfty$ this becomes just
$$frac13+frac15+frac17.$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    After given



    $$sum_{n=1}^{infty} dfrac{1}{(2n+1)(2n+7)}=frac1{36}sum_{n=1}^{infty} dfrac{1}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



    set function



    $$f(x)=frac1{36}sum_{n=1}^{infty} dfrac{x^{frac{n}{3}+frac{7}{6}}}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



    then take the second derivative of function $f(x)$, which is



    $$f''(x)=frac1{36}sum_{n=1}^{infty} x^{frac{n}{3}-frac{5}{6}}$$



    and it is easy to find this series is equal to



    $$f''(x)=frac{1}{36x^{1/2}(1-x^{1/3})}$$



    also, notice that you also get the boundary $f'(0)=0$ and $f(0)=0$, then you can do the integral twice to find the original $f(x)$, which is



    $$f(x)=frac{x^{1/6}}{6}(1+frac{1}{3}x^{1/3}+frac{1}{5}x^{2/3}-frac{6}{7}x)+frac{1-x}{12}lnleft(frac{1+x^{1/6}}{1-x^{1/6}}right)$$



    and take the limitation for $xto1$ which is the result



    $$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}= lim_{x to 1} f(x)=frac{71}{630}$$



    Actually, the general function method is much more complicated than fraction splitting.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      Hint: Let's look at the $100$th partial sum. It's good to get some concreteness.



      $$frac{1}{6}left(frac{1}{3}+frac{1}{5}+cdots+frac{1}{201}right)-frac{1}{6}left(frac{1}{9}+cdots+frac{1}{205}+frac{1}{207}right).$$



      We have a bunch of terms that are repeated: $frac{1}{9}+cdots+frac{1}{201}$ exists in each bracketed portion, so we can simply cancel all of them out to get



      $$frac{1}{6}left(frac{1}{3}+frac{1}{5}+frac{1}{7}-frac{1}{203}-frac{1}{205}-frac{1}{207}right).$$



      Can you see how to use this line of reasoning to get the answer?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        (+1) Nice hint!
        $endgroup$
        – José Carlos Santos
        Dec 8 '18 at 15:40
















      7












      $begingroup$

      Hint: Let's look at the $100$th partial sum. It's good to get some concreteness.



      $$frac{1}{6}left(frac{1}{3}+frac{1}{5}+cdots+frac{1}{201}right)-frac{1}{6}left(frac{1}{9}+cdots+frac{1}{205}+frac{1}{207}right).$$



      We have a bunch of terms that are repeated: $frac{1}{9}+cdots+frac{1}{201}$ exists in each bracketed portion, so we can simply cancel all of them out to get



      $$frac{1}{6}left(frac{1}{3}+frac{1}{5}+frac{1}{7}-frac{1}{203}-frac{1}{205}-frac{1}{207}right).$$



      Can you see how to use this line of reasoning to get the answer?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        (+1) Nice hint!
        $endgroup$
        – José Carlos Santos
        Dec 8 '18 at 15:40














      7












      7








      7





      $begingroup$

      Hint: Let's look at the $100$th partial sum. It's good to get some concreteness.



      $$frac{1}{6}left(frac{1}{3}+frac{1}{5}+cdots+frac{1}{201}right)-frac{1}{6}left(frac{1}{9}+cdots+frac{1}{205}+frac{1}{207}right).$$



      We have a bunch of terms that are repeated: $frac{1}{9}+cdots+frac{1}{201}$ exists in each bracketed portion, so we can simply cancel all of them out to get



      $$frac{1}{6}left(frac{1}{3}+frac{1}{5}+frac{1}{7}-frac{1}{203}-frac{1}{205}-frac{1}{207}right).$$



      Can you see how to use this line of reasoning to get the answer?






      share|cite|improve this answer









      $endgroup$



      Hint: Let's look at the $100$th partial sum. It's good to get some concreteness.



      $$frac{1}{6}left(frac{1}{3}+frac{1}{5}+cdots+frac{1}{201}right)-frac{1}{6}left(frac{1}{9}+cdots+frac{1}{205}+frac{1}{207}right).$$



      We have a bunch of terms that are repeated: $frac{1}{9}+cdots+frac{1}{201}$ exists in each bracketed portion, so we can simply cancel all of them out to get



      $$frac{1}{6}left(frac{1}{3}+frac{1}{5}+frac{1}{7}-frac{1}{203}-frac{1}{205}-frac{1}{207}right).$$



      Can you see how to use this line of reasoning to get the answer?







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 8 '18 at 15:38









      Carl SchildkrautCarl Schildkraut

      11.3k11441




      11.3k11441












      • $begingroup$
        (+1) Nice hint!
        $endgroup$
        – José Carlos Santos
        Dec 8 '18 at 15:40


















      • $begingroup$
        (+1) Nice hint!
        $endgroup$
        – José Carlos Santos
        Dec 8 '18 at 15:40
















      $begingroup$
      (+1) Nice hint!
      $endgroup$
      – José Carlos Santos
      Dec 8 '18 at 15:40




      $begingroup$
      (+1) Nice hint!
      $endgroup$
      – José Carlos Santos
      Dec 8 '18 at 15:40











      3












      $begingroup$

      Check that
      $$left(frac13+frac15+frac17+frac19+frac1{11}+cdots+frac1{2N+1}right)-$$
      $$-left(frac19+frac1{11}+cdots+frac1{2N+1}+frac1{2N+3}+frac1{2N+5}+frac1{2N+7}right)=$$
      $$=frac13+frac15+frac17-frac1{2N+3}-frac1{2N+5}-frac1{2N+7}.$$



      And if you take the limit as $Ntoinfty$ this becomes just
      $$frac13+frac15+frac17.$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Check that
        $$left(frac13+frac15+frac17+frac19+frac1{11}+cdots+frac1{2N+1}right)-$$
        $$-left(frac19+frac1{11}+cdots+frac1{2N+1}+frac1{2N+3}+frac1{2N+5}+frac1{2N+7}right)=$$
        $$=frac13+frac15+frac17-frac1{2N+3}-frac1{2N+5}-frac1{2N+7}.$$



        And if you take the limit as $Ntoinfty$ this becomes just
        $$frac13+frac15+frac17.$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Check that
          $$left(frac13+frac15+frac17+frac19+frac1{11}+cdots+frac1{2N+1}right)-$$
          $$-left(frac19+frac1{11}+cdots+frac1{2N+1}+frac1{2N+3}+frac1{2N+5}+frac1{2N+7}right)=$$
          $$=frac13+frac15+frac17-frac1{2N+3}-frac1{2N+5}-frac1{2N+7}.$$



          And if you take the limit as $Ntoinfty$ this becomes just
          $$frac13+frac15+frac17.$$






          share|cite|improve this answer









          $endgroup$



          Check that
          $$left(frac13+frac15+frac17+frac19+frac1{11}+cdots+frac1{2N+1}right)-$$
          $$-left(frac19+frac1{11}+cdots+frac1{2N+1}+frac1{2N+3}+frac1{2N+5}+frac1{2N+7}right)=$$
          $$=frac13+frac15+frac17-frac1{2N+3}-frac1{2N+5}-frac1{2N+7}.$$



          And if you take the limit as $Ntoinfty$ this becomes just
          $$frac13+frac15+frac17.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 15:40









          Alejandro Nasif SalumAlejandro Nasif Salum

          4,765118




          4,765118























              2












              $begingroup$

              After given



              $$sum_{n=1}^{infty} dfrac{1}{(2n+1)(2n+7)}=frac1{36}sum_{n=1}^{infty} dfrac{1}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



              set function



              $$f(x)=frac1{36}sum_{n=1}^{infty} dfrac{x^{frac{n}{3}+frac{7}{6}}}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



              then take the second derivative of function $f(x)$, which is



              $$f''(x)=frac1{36}sum_{n=1}^{infty} x^{frac{n}{3}-frac{5}{6}}$$



              and it is easy to find this series is equal to



              $$f''(x)=frac{1}{36x^{1/2}(1-x^{1/3})}$$



              also, notice that you also get the boundary $f'(0)=0$ and $f(0)=0$, then you can do the integral twice to find the original $f(x)$, which is



              $$f(x)=frac{x^{1/6}}{6}(1+frac{1}{3}x^{1/3}+frac{1}{5}x^{2/3}-frac{6}{7}x)+frac{1-x}{12}lnleft(frac{1+x^{1/6}}{1-x^{1/6}}right)$$



              and take the limitation for $xto1$ which is the result



              $$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}= lim_{x to 1} f(x)=frac{71}{630}$$



              Actually, the general function method is much more complicated than fraction splitting.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                After given



                $$sum_{n=1}^{infty} dfrac{1}{(2n+1)(2n+7)}=frac1{36}sum_{n=1}^{infty} dfrac{1}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



                set function



                $$f(x)=frac1{36}sum_{n=1}^{infty} dfrac{x^{frac{n}{3}+frac{7}{6}}}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



                then take the second derivative of function $f(x)$, which is



                $$f''(x)=frac1{36}sum_{n=1}^{infty} x^{frac{n}{3}-frac{5}{6}}$$



                and it is easy to find this series is equal to



                $$f''(x)=frac{1}{36x^{1/2}(1-x^{1/3})}$$



                also, notice that you also get the boundary $f'(0)=0$ and $f(0)=0$, then you can do the integral twice to find the original $f(x)$, which is



                $$f(x)=frac{x^{1/6}}{6}(1+frac{1}{3}x^{1/3}+frac{1}{5}x^{2/3}-frac{6}{7}x)+frac{1-x}{12}lnleft(frac{1+x^{1/6}}{1-x^{1/6}}right)$$



                and take the limitation for $xto1$ which is the result



                $$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}= lim_{x to 1} f(x)=frac{71}{630}$$



                Actually, the general function method is much more complicated than fraction splitting.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  After given



                  $$sum_{n=1}^{infty} dfrac{1}{(2n+1)(2n+7)}=frac1{36}sum_{n=1}^{infty} dfrac{1}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



                  set function



                  $$f(x)=frac1{36}sum_{n=1}^{infty} dfrac{x^{frac{n}{3}+frac{7}{6}}}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



                  then take the second derivative of function $f(x)$, which is



                  $$f''(x)=frac1{36}sum_{n=1}^{infty} x^{frac{n}{3}-frac{5}{6}}$$



                  and it is easy to find this series is equal to



                  $$f''(x)=frac{1}{36x^{1/2}(1-x^{1/3})}$$



                  also, notice that you also get the boundary $f'(0)=0$ and $f(0)=0$, then you can do the integral twice to find the original $f(x)$, which is



                  $$f(x)=frac{x^{1/6}}{6}(1+frac{1}{3}x^{1/3}+frac{1}{5}x^{2/3}-frac{6}{7}x)+frac{1-x}{12}lnleft(frac{1+x^{1/6}}{1-x^{1/6}}right)$$



                  and take the limitation for $xto1$ which is the result



                  $$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}= lim_{x to 1} f(x)=frac{71}{630}$$



                  Actually, the general function method is much more complicated than fraction splitting.






                  share|cite|improve this answer











                  $endgroup$



                  After given



                  $$sum_{n=1}^{infty} dfrac{1}{(2n+1)(2n+7)}=frac1{36}sum_{n=1}^{infty} dfrac{1}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



                  set function



                  $$f(x)=frac1{36}sum_{n=1}^{infty} dfrac{x^{frac{n}{3}+frac{7}{6}}}{(frac{n}{3}+frac{1}{6})(frac{n}{3}+frac{7}{6})}$$



                  then take the second derivative of function $f(x)$, which is



                  $$f''(x)=frac1{36}sum_{n=1}^{infty} x^{frac{n}{3}-frac{5}{6}}$$



                  and it is easy to find this series is equal to



                  $$f''(x)=frac{1}{36x^{1/2}(1-x^{1/3})}$$



                  also, notice that you also get the boundary $f'(0)=0$ and $f(0)=0$, then you can do the integral twice to find the original $f(x)$, which is



                  $$f(x)=frac{x^{1/6}}{6}(1+frac{1}{3}x^{1/3}+frac{1}{5}x^{2/3}-frac{6}{7}x)+frac{1-x}{12}lnleft(frac{1+x^{1/6}}{1-x^{1/6}}right)$$



                  and take the limitation for $xto1$ which is the result



                  $$sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}= lim_{x to 1} f(x)=frac{71}{630}$$



                  Actually, the general function method is much more complicated than fraction splitting.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 8 '18 at 22:34

























                  answered Dec 8 '18 at 16:25









                  NanayajitzukiNanayajitzuki

                  3185




                  3185






























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