$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}}$












6












$begingroup$


$$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}}$$



We know that : $arctan{x} - arctan{y} = arctan{frac{x-y}{1+xy}}$ for every $ xy > 1 $



I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.



Edit: I am very sorry on the denominator it should have been $n^2+n+4$ not $n^2+2n+4$



Similarly here :



$$sumlimits_{n=1}^{infty}arctan{frac{8n}{n^4-2n^2+5}}$$



The result should be $ arctan 2 $ on the first one and $ pi/2 + arctan2 $ on the second one.






calculus sequences-and-series trigonometry telescopic-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use $b=a-2$ and solve for $a$, the quadratic $a(a-2)=n^2+2n+3$
    $endgroup$
    – Claude Leibovici
    Jan 28 at 9:48






  • 1




    $begingroup$
    If $a = 2+b$, then $(2+b)b = n^2 + 2n + 3$. Solve.
    $endgroup$
    – Euler....IS_ALIVE
    Jan 28 at 9:49
















6












$begingroup$


$$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}}$$



We know that : $arctan{x} - arctan{y} = arctan{frac{x-y}{1+xy}}$ for every $ xy > 1 $



I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.



Edit: I am very sorry on the denominator it should have been $n^2+n+4$ not $n^2+2n+4$



Similarly here :



$$sumlimits_{n=1}^{infty}arctan{frac{8n}{n^4-2n^2+5}}$$



The result should be $ arctan 2 $ on the first one and $ pi/2 + arctan2 $ on the second one.






calculus sequences-and-series trigonometry telescopic-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use $b=a-2$ and solve for $a$, the quadratic $a(a-2)=n^2+2n+3$
    $endgroup$
    – Claude Leibovici
    Jan 28 at 9:48






  • 1




    $begingroup$
    If $a = 2+b$, then $(2+b)b = n^2 + 2n + 3$. Solve.
    $endgroup$
    – Euler....IS_ALIVE
    Jan 28 at 9:49














6












6








6


1



$begingroup$


$$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}}$$



We know that : $arctan{x} - arctan{y} = arctan{frac{x-y}{1+xy}}$ for every $ xy > 1 $



I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.



Edit: I am very sorry on the denominator it should have been $n^2+n+4$ not $n^2+2n+4$



Similarly here :



$$sumlimits_{n=1}^{infty}arctan{frac{8n}{n^4-2n^2+5}}$$



The result should be $ arctan 2 $ on the first one and $ pi/2 + arctan2 $ on the second one.






calculus sequences-and-series trigonometry telescopic-series







share|cite|improve this question











$endgroup$




$$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}}$$



We know that : $arctan{x} - arctan{y} = arctan{frac{x-y}{1+xy}}$ for every $ xy > 1 $



I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.



Edit: I am very sorry on the denominator it should have been $n^2+n+4$ not $n^2+2n+4$



Similarly here :



$$sumlimits_{n=1}^{infty}arctan{frac{8n}{n^4-2n^2+5}}$$



The result should be $ arctan 2 $ on the first one and $ pi/2 + arctan2 $ on the second one.






calculus sequences-and-series trigonometry telescopic-series




calculus sequences-and-series trigonometry telescopic-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 11 at 10:01









lab bhattacharjee

228k15158278




228k15158278










asked Jan 28 at 9:01









SADBOYSSADBOYS

51119




51119








  • 1




    $begingroup$
    Use $b=a-2$ and solve for $a$, the quadratic $a(a-2)=n^2+2n+3$
    $endgroup$
    – Claude Leibovici
    Jan 28 at 9:48






  • 1




    $begingroup$
    If $a = 2+b$, then $(2+b)b = n^2 + 2n + 3$. Solve.
    $endgroup$
    – Euler....IS_ALIVE
    Jan 28 at 9:49














  • 1




    $begingroup$
    Use $b=a-2$ and solve for $a$, the quadratic $a(a-2)=n^2+2n+3$
    $endgroup$
    – Claude Leibovici
    Jan 28 at 9:48






  • 1




    $begingroup$
    If $a = 2+b$, then $(2+b)b = n^2 + 2n + 3$. Solve.
    $endgroup$
    – Euler....IS_ALIVE
    Jan 28 at 9:49








1




1




$begingroup$
Use $b=a-2$ and solve for $a$, the quadratic $a(a-2)=n^2+2n+3$
$endgroup$
– Claude Leibovici
Jan 28 at 9:48




$begingroup$
Use $b=a-2$ and solve for $a$, the quadratic $a(a-2)=n^2+2n+3$
$endgroup$
– Claude Leibovici
Jan 28 at 9:48




1




1




$begingroup$
If $a = 2+b$, then $(2+b)b = n^2 + 2n + 3$. Solve.
$endgroup$
– Euler....IS_ALIVE
Jan 28 at 9:49




$begingroup$
If $a = 2+b$, then $(2+b)b = n^2 + 2n + 3$. Solve.
$endgroup$
– Euler....IS_ALIVE
Jan 28 at 9:49










2 Answers
2






active

oldest

votes


















8












$begingroup$

Hint:



$$dfrac2{n^2+n+4}=dfrac{dfrac12}{1+dfrac{n(n+1)}4}=dfrac{dfrac{n+1}2-dfrac n2}{1+dfrac n2cdotdfrac{n+1}2}$$



For the second, $$dfrac{8n}{n^4-2n^2+5}=dfrac{dfrac{(n+1)^2}2-dfrac{(n-1)^2}2}{dfrac{(n+1)^2}2cdotdfrac{(n-1)^2}2+1}$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    See also: math.stackexchange.com/questions/193001/… math.stackexchange.com/questions/2789725/…
    $endgroup$
    – lab bhattacharjee
    Jan 28 at 11:58










  • $begingroup$
    Can you explain in more details how the expressions for $a$ and $b$ in both cases were found?
    $endgroup$
    – user
    Feb 19 at 12:27










  • $begingroup$
    @user, Where are $a,b$ in the current version ?
    $endgroup$
    – lab bhattacharjee
    Feb 19 at 12:53










  • $begingroup$
    @user, math.stackexchange.com/questions/3107476/…
    $endgroup$
    – lab bhattacharjee
    Feb 19 at 13:09










  • $begingroup$
    Thank you very much! Now I see. Do I understand correctly that in the second case you used $f(n)=arctan(an^2+bn+c)$ as a try function?
    $endgroup$
    – user
    Feb 19 at 13:20



















1












$begingroup$

$$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}} = sumlimits_{n=1}^{infty}arctan{frac{2(n+1)-2n}{n(n+1)(1+frac{4}{n(n+1)})}} =
sumlimits_{n=1}^{infty}arctan{frac{frac{2}{n}-frac{2}{n+1}}{1+frac{2}
{n}cdotfrac{2}{n+1}}} =sumlimits_{n=1}^{infty}arctan{frac{2}{n}-arctan{frac{2}{n+1}}} =arctan2$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    Hint:



    $$dfrac2{n^2+n+4}=dfrac{dfrac12}{1+dfrac{n(n+1)}4}=dfrac{dfrac{n+1}2-dfrac n2}{1+dfrac n2cdotdfrac{n+1}2}$$



    For the second, $$dfrac{8n}{n^4-2n^2+5}=dfrac{dfrac{(n+1)^2}2-dfrac{(n-1)^2}2}{dfrac{(n+1)^2}2cdotdfrac{(n-1)^2}2+1}$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      See also: math.stackexchange.com/questions/193001/… math.stackexchange.com/questions/2789725/…
      $endgroup$
      – lab bhattacharjee
      Jan 28 at 11:58










    • $begingroup$
      Can you explain in more details how the expressions for $a$ and $b$ in both cases were found?
      $endgroup$
      – user
      Feb 19 at 12:27










    • $begingroup$
      @user, Where are $a,b$ in the current version ?
      $endgroup$
      – lab bhattacharjee
      Feb 19 at 12:53










    • $begingroup$
      @user, math.stackexchange.com/questions/3107476/…
      $endgroup$
      – lab bhattacharjee
      Feb 19 at 13:09










    • $begingroup$
      Thank you very much! Now I see. Do I understand correctly that in the second case you used $f(n)=arctan(an^2+bn+c)$ as a try function?
      $endgroup$
      – user
      Feb 19 at 13:20
















    8












    $begingroup$

    Hint:



    $$dfrac2{n^2+n+4}=dfrac{dfrac12}{1+dfrac{n(n+1)}4}=dfrac{dfrac{n+1}2-dfrac n2}{1+dfrac n2cdotdfrac{n+1}2}$$



    For the second, $$dfrac{8n}{n^4-2n^2+5}=dfrac{dfrac{(n+1)^2}2-dfrac{(n-1)^2}2}{dfrac{(n+1)^2}2cdotdfrac{(n-1)^2}2+1}$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      See also: math.stackexchange.com/questions/193001/… math.stackexchange.com/questions/2789725/…
      $endgroup$
      – lab bhattacharjee
      Jan 28 at 11:58










    • $begingroup$
      Can you explain in more details how the expressions for $a$ and $b$ in both cases were found?
      $endgroup$
      – user
      Feb 19 at 12:27










    • $begingroup$
      @user, Where are $a,b$ in the current version ?
      $endgroup$
      – lab bhattacharjee
      Feb 19 at 12:53










    • $begingroup$
      @user, math.stackexchange.com/questions/3107476/…
      $endgroup$
      – lab bhattacharjee
      Feb 19 at 13:09










    • $begingroup$
      Thank you very much! Now I see. Do I understand correctly that in the second case you used $f(n)=arctan(an^2+bn+c)$ as a try function?
      $endgroup$
      – user
      Feb 19 at 13:20














    8












    8








    8





    $begingroup$

    Hint:



    $$dfrac2{n^2+n+4}=dfrac{dfrac12}{1+dfrac{n(n+1)}4}=dfrac{dfrac{n+1}2-dfrac n2}{1+dfrac n2cdotdfrac{n+1}2}$$



    For the second, $$dfrac{8n}{n^4-2n^2+5}=dfrac{dfrac{(n+1)^2}2-dfrac{(n-1)^2}2}{dfrac{(n+1)^2}2cdotdfrac{(n-1)^2}2+1}$$






    share|cite|improve this answer











    $endgroup$



    Hint:



    $$dfrac2{n^2+n+4}=dfrac{dfrac12}{1+dfrac{n(n+1)}4}=dfrac{dfrac{n+1}2-dfrac n2}{1+dfrac n2cdotdfrac{n+1}2}$$



    For the second, $$dfrac{8n}{n^4-2n^2+5}=dfrac{dfrac{(n+1)^2}2-dfrac{(n-1)^2}2}{dfrac{(n+1)^2}2cdotdfrac{(n-1)^2}2+1}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Feb 19 at 10:33

























    answered Jan 28 at 11:36









    lab bhattacharjeelab bhattacharjee

    228k15158278




    228k15158278








    • 1




      $begingroup$
      See also: math.stackexchange.com/questions/193001/… math.stackexchange.com/questions/2789725/…
      $endgroup$
      – lab bhattacharjee
      Jan 28 at 11:58










    • $begingroup$
      Can you explain in more details how the expressions for $a$ and $b$ in both cases were found?
      $endgroup$
      – user
      Feb 19 at 12:27










    • $begingroup$
      @user, Where are $a,b$ in the current version ?
      $endgroup$
      – lab bhattacharjee
      Feb 19 at 12:53










    • $begingroup$
      @user, math.stackexchange.com/questions/3107476/…
      $endgroup$
      – lab bhattacharjee
      Feb 19 at 13:09










    • $begingroup$
      Thank you very much! Now I see. Do I understand correctly that in the second case you used $f(n)=arctan(an^2+bn+c)$ as a try function?
      $endgroup$
      – user
      Feb 19 at 13:20














    • 1




      $begingroup$
      See also: math.stackexchange.com/questions/193001/… math.stackexchange.com/questions/2789725/…
      $endgroup$
      – lab bhattacharjee
      Jan 28 at 11:58










    • $begingroup$
      Can you explain in more details how the expressions for $a$ and $b$ in both cases were found?
      $endgroup$
      – user
      Feb 19 at 12:27










    • $begingroup$
      @user, Where are $a,b$ in the current version ?
      $endgroup$
      – lab bhattacharjee
      Feb 19 at 12:53










    • $begingroup$
      @user, math.stackexchange.com/questions/3107476/…
      $endgroup$
      – lab bhattacharjee
      Feb 19 at 13:09










    • $begingroup$
      Thank you very much! Now I see. Do I understand correctly that in the second case you used $f(n)=arctan(an^2+bn+c)$ as a try function?
      $endgroup$
      – user
      Feb 19 at 13:20








    1




    1




    $begingroup$
    See also: math.stackexchange.com/questions/193001/… math.stackexchange.com/questions/2789725/…
    $endgroup$
    – lab bhattacharjee
    Jan 28 at 11:58




    $begingroup$
    See also: math.stackexchange.com/questions/193001/… math.stackexchange.com/questions/2789725/…
    $endgroup$
    – lab bhattacharjee
    Jan 28 at 11:58












    $begingroup$
    Can you explain in more details how the expressions for $a$ and $b$ in both cases were found?
    $endgroup$
    – user
    Feb 19 at 12:27




    $begingroup$
    Can you explain in more details how the expressions for $a$ and $b$ in both cases were found?
    $endgroup$
    – user
    Feb 19 at 12:27












    $begingroup$
    @user, Where are $a,b$ in the current version ?
    $endgroup$
    – lab bhattacharjee
    Feb 19 at 12:53




    $begingroup$
    @user, Where are $a,b$ in the current version ?
    $endgroup$
    – lab bhattacharjee
    Feb 19 at 12:53












    $begingroup$
    @user, math.stackexchange.com/questions/3107476/…
    $endgroup$
    – lab bhattacharjee
    Feb 19 at 13:09




    $begingroup$
    @user, math.stackexchange.com/questions/3107476/…
    $endgroup$
    – lab bhattacharjee
    Feb 19 at 13:09












    $begingroup$
    Thank you very much! Now I see. Do I understand correctly that in the second case you used $f(n)=arctan(an^2+bn+c)$ as a try function?
    $endgroup$
    – user
    Feb 19 at 13:20




    $begingroup$
    Thank you very much! Now I see. Do I understand correctly that in the second case you used $f(n)=arctan(an^2+bn+c)$ as a try function?
    $endgroup$
    – user
    Feb 19 at 13:20











    1












    $begingroup$

    $$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}} = sumlimits_{n=1}^{infty}arctan{frac{2(n+1)-2n}{n(n+1)(1+frac{4}{n(n+1)})}} =
    sumlimits_{n=1}^{infty}arctan{frac{frac{2}{n}-frac{2}{n+1}}{1+frac{2}
    {n}cdotfrac{2}{n+1}}} =sumlimits_{n=1}^{infty}arctan{frac{2}{n}-arctan{frac{2}{n+1}}} =arctan2$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}} = sumlimits_{n=1}^{infty}arctan{frac{2(n+1)-2n}{n(n+1)(1+frac{4}{n(n+1)})}} =
      sumlimits_{n=1}^{infty}arctan{frac{frac{2}{n}-frac{2}{n+1}}{1+frac{2}
      {n}cdotfrac{2}{n+1}}} =sumlimits_{n=1}^{infty}arctan{frac{2}{n}-arctan{frac{2}{n+1}}} =arctan2$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}} = sumlimits_{n=1}^{infty}arctan{frac{2(n+1)-2n}{n(n+1)(1+frac{4}{n(n+1)})}} =
        sumlimits_{n=1}^{infty}arctan{frac{frac{2}{n}-frac{2}{n+1}}{1+frac{2}
        {n}cdotfrac{2}{n+1}}} =sumlimits_{n=1}^{infty}arctan{frac{2}{n}-arctan{frac{2}{n+1}}} =arctan2$$






        share|cite|improve this answer









        $endgroup$



        $$sumlimits_{n=1}^{infty}arctan{frac{2}{n^2+n+4}} = sumlimits_{n=1}^{infty}arctan{frac{2(n+1)-2n}{n(n+1)(1+frac{4}{n(n+1)})}} =
        sumlimits_{n=1}^{infty}arctan{frac{frac{2}{n}-frac{2}{n+1}}{1+frac{2}
        {n}cdotfrac{2}{n+1}}} =sumlimits_{n=1}^{infty}arctan{frac{2}{n}-arctan{frac{2}{n+1}}} =arctan2$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 11:41









        SADBOYSSADBOYS

        51119




        51119






























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