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







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












$begingroup$


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











    $endgroup$













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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      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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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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                      Residenzschloss Arolsen

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                      Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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