What is the value of $frac11+frac13-frac15-frac17+frac19+frac111-dots$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Find the sum of $sum 1/(k^2 - a^2)$ when $0<a<1$Find the limit of $sum frac1log^n(n)$Calculating the infinite series $1-frac13+frac15-frac17+frac19-frac111cdots$Another SummationNumerical convergence depending on summation orderSign flip for every TWO terms in a sum? (Instead of one)Show that $sum_n=0^infty |a_n|<infty.$A series whose terms are the products of terms of a geometric and a power seriesCheck if the series $sum_n=1^inftyfrac(-1)^fracn(n-1)2sqrt n$ convergesConvergence of $1+frac13-frac12+frac15+frac17-frac14+frac19+frac111-frac16+ldots$What is the sign and exact value of this sum: $sum_n=0^+infty(-1)^n tan(frac1n!)$?

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What is the value of $frac11+frac13-frac15-frac17+frac19+frac111-dots$?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Find the sum of $sum 1/(k^2 - a^2)$ when $0<a<1$Find the limit of $sum frac1log^n(n)$Calculating the infinite series $1-frac13+frac15-frac17+frac19-frac111cdots$Another SummationNumerical convergence depending on summation orderSign flip for every TWO terms in a sum? (Instead of one)Show that $sum_n=0^infty |a_n|<infty.$A series whose terms are the products of terms of a geometric and a power seriesCheck if the series $sum_n=1^inftyfrac(-1)^fracn(n-1)2sqrt n$ convergesConvergence of $1+frac13-frac12+frac15+frac17-frac14+frac19+frac111-frac16+ldots$What is the sign and exact value of this sum: $sum_n=0^+infty(-1)^n tan(frac1n!)$?










6












$begingroup$


The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.



The series $sum_k=1^infty frac(-1)^frack^2+k+222k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.




What is the convergence value, explicitly, of the second series?




The first summation is noted above, because it might be a useful information to evaluate the second summation.






sequences-and-series summation integers pi







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Wolfram says the result is $fracsqrt2pi4$
    $endgroup$
    – Peter Foreman
    Apr 17 at 10:28










  • $begingroup$
    @PeterForeman yes, but how to approach that value?
    $endgroup$
    – Hussain-Alqatari
    Apr 17 at 10:40















6












$begingroup$


The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.



The series $sum_k=1^infty frac(-1)^frack^2+k+222k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.




What is the convergence value, explicitly, of the second series?




The first summation is noted above, because it might be a useful information to evaluate the second summation.






sequences-and-series summation integers pi







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Wolfram says the result is $fracsqrt2pi4$
    $endgroup$
    – Peter Foreman
    Apr 17 at 10:28










  • $begingroup$
    @PeterForeman yes, but how to approach that value?
    $endgroup$
    – Hussain-Alqatari
    Apr 17 at 10:40













6












6








6


5



$begingroup$


The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.



The series $sum_k=1^infty frac(-1)^frack^2+k+222k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.




What is the convergence value, explicitly, of the second series?




The first summation is noted above, because it might be a useful information to evaluate the second summation.






sequences-and-series summation integers pi







share|cite|improve this question











$endgroup$




The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.



The series $sum_k=1^infty frac(-1)^frack^2+k+222k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.




What is the convergence value, explicitly, of the second series?




The first summation is noted above, because it might be a useful information to evaluate the second summation.






sequences-and-series summation integers pi




sequences-and-series summation integers pi






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 17 at 15:09









user21820

40.4k544163




40.4k544163










asked Apr 17 at 10:08









Hussain-AlqatariHussain-Alqatari

3587




3587







  • 1




    $begingroup$
    Wolfram says the result is $fracsqrt2pi4$
    $endgroup$
    – Peter Foreman
    Apr 17 at 10:28










  • $begingroup$
    @PeterForeman yes, but how to approach that value?
    $endgroup$
    – Hussain-Alqatari
    Apr 17 at 10:40












  • 1




    $begingroup$
    Wolfram says the result is $fracsqrt2pi4$
    $endgroup$
    – Peter Foreman
    Apr 17 at 10:28










  • $begingroup$
    @PeterForeman yes, but how to approach that value?
    $endgroup$
    – Hussain-Alqatari
    Apr 17 at 10:40







1




1




$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28




$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28












$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40




$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40










3 Answers
3






active

oldest

votes


















12












$begingroup$

Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$



The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.






share|cite|improve this answer









$endgroup$








  • 2




    $begingroup$
    How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
    $endgroup$
    – Hussain-Alqatari
    Apr 17 at 12:56











  • $begingroup$
    Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
    $endgroup$
    – Michael Seifert
    Apr 17 at 14:11


















5












$begingroup$

Hint:



$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$



$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$



$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$



$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$



then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    $$
    beginalign
    &sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
    &=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
    &=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
    &=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
    &=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
    &=fracpisqrt24tag5
    endalign
    $$
    Explanation:
    $(1)$: separate two absolutely convergent series
    $(2)$: each series can be written as a sum over $mathbbZ$
    $(3)$: factor $frac18$ out of each series
    $(4)$: apply $(7)$ from this answer
    $(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
      $endgroup$
      – robjohn♦
      Apr 18 at 12:05












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    12












    $begingroup$

    Let $$beginalign
    S_1&=1-frac15+frac19-1over13+dots\
    S_2&=frac13-frac17+1over11-1over15+dots
    endalign$$     so that the sum we seek is $S_1+S_2.$
    T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
    and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
    By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$



    The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.






    share|cite|improve this answer









    $endgroup$








    • 2




      $begingroup$
      How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
      $endgroup$
      – Hussain-Alqatari
      Apr 17 at 12:56











    • $begingroup$
      Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
      $endgroup$
      – Michael Seifert
      Apr 17 at 14:11















    12












    $begingroup$

    Let $$beginalign
    S_1&=1-frac15+frac19-1over13+dots\
    S_2&=frac13-frac17+1over11-1over15+dots
    endalign$$     so that the sum we seek is $S_1+S_2.$
    T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
    and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
    By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$



    The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.






    share|cite|improve this answer









    $endgroup$








    • 2




      $begingroup$
      How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
      $endgroup$
      – Hussain-Alqatari
      Apr 17 at 12:56











    • $begingroup$
      Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
      $endgroup$
      – Michael Seifert
      Apr 17 at 14:11













    12












    12








    12





    $begingroup$

    Let $$beginalign
    S_1&=1-frac15+frac19-1over13+dots\
    S_2&=frac13-frac17+1over11-1over15+dots
    endalign$$     so that the sum we seek is $S_1+S_2.$
    T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
    and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
    By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$



    The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.






    share|cite|improve this answer









    $endgroup$



    Let $$beginalign
    S_1&=1-frac15+frac19-1over13+dots\
    S_2&=frac13-frac17+1over11-1over15+dots
    endalign$$     so that the sum we seek is $S_1+S_2.$
    T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
    and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
    By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$



    The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 17 at 11:03









    saulspatzsaulspatz

    17.8k31536




    17.8k31536







    • 2




      $begingroup$
      How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
      $endgroup$
      – Hussain-Alqatari
      Apr 17 at 12:56











    • $begingroup$
      Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
      $endgroup$
      – Michael Seifert
      Apr 17 at 14:11












    • 2




      $begingroup$
      How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
      $endgroup$
      – Hussain-Alqatari
      Apr 17 at 12:56











    • $begingroup$
      Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
      $endgroup$
      – Michael Seifert
      Apr 17 at 14:11







    2




    2




    $begingroup$
    How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
    $endgroup$
    – Hussain-Alqatari
    Apr 17 at 12:56





    $begingroup$
    How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
    $endgroup$
    – Hussain-Alqatari
    Apr 17 at 12:56













    $begingroup$
    Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
    $endgroup$
    – Michael Seifert
    Apr 17 at 14:11




    $begingroup$
    Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
    $endgroup$
    – Michael Seifert
    Apr 17 at 14:11











    5












    $begingroup$

    Hint:



    $$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$



    $$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$



    $$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$



    $$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
    $$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$



    then use
    $$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$






    share|cite|improve this answer











    $endgroup$

















      5












      $begingroup$

      Hint:



      $$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$



      $$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$



      $$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$



      $$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
      $$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$



      then use
      $$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$






      share|cite|improve this answer











      $endgroup$















        5












        5








        5





        $begingroup$

        Hint:



        $$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$



        $$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$



        $$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$



        $$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
        $$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$



        then use
        $$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$






        share|cite|improve this answer











        $endgroup$



        Hint:



        $$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$



        $$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$



        $$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$



        $$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
        $$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$



        then use
        $$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 17 at 14:52

























        answered Apr 17 at 12:30









        E.H.EE.H.E

        17k11969




        17k11969





















            1












            $begingroup$

            $$
            beginalign
            &sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
            &=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
            &=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
            &=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
            &=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
            &=fracpisqrt24tag5
            endalign
            $$
            Explanation:
            $(1)$: separate two absolutely convergent series
            $(2)$: each series can be written as a sum over $mathbbZ$
            $(3)$: factor $frac18$ out of each series
            $(4)$: apply $(7)$ from this answer
            $(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
              $endgroup$
              – robjohn♦
              Apr 18 at 12:05
















            1












            $begingroup$

            $$
            beginalign
            &sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
            &=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
            &=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
            &=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
            &=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
            &=fracpisqrt24tag5
            endalign
            $$
            Explanation:
            $(1)$: separate two absolutely convergent series
            $(2)$: each series can be written as a sum over $mathbbZ$
            $(3)$: factor $frac18$ out of each series
            $(4)$: apply $(7)$ from this answer
            $(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
              $endgroup$
              – robjohn♦
              Apr 18 at 12:05














            1












            1








            1





            $begingroup$

            $$
            beginalign
            &sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
            &=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
            &=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
            &=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
            &=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
            &=fracpisqrt24tag5
            endalign
            $$
            Explanation:
            $(1)$: separate two absolutely convergent series
            $(2)$: each series can be written as a sum over $mathbbZ$
            $(3)$: factor $frac18$ out of each series
            $(4)$: apply $(7)$ from this answer
            $(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$






            share|cite|improve this answer











            $endgroup$



            $$
            beginalign
            &sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
            &=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
            &=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
            &=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
            &=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
            &=fracpisqrt24tag5
            endalign
            $$
            Explanation:
            $(1)$: separate two absolutely convergent series
            $(2)$: each series can be written as a sum over $mathbbZ$
            $(3)$: factor $frac18$ out of each series
            $(4)$: apply $(7)$ from this answer
            $(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 17 at 17:49

























            answered Apr 17 at 17:33









            robjohn♦robjohn

            271k27316643




            271k27316643











            • $begingroup$
              For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
              $endgroup$
              – robjohn♦
              Apr 18 at 12:05

















            • $begingroup$
              For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
              $endgroup$
              – robjohn♦
              Apr 18 at 12:05
















            $begingroup$
            For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
            $endgroup$
            – robjohn♦
            Apr 18 at 12:05





            $begingroup$
            For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
            $endgroup$
            – robjohn♦
            Apr 18 at 12:05


















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