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Kate
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I can show that $\sum^\infty_{k=1}{{1}\over{k^2}} = {{\pi^2}\over{6}}$ without to much hassle just using two representations of ${{sin(x)}\over{x}}$ but I cant find anythingany proof nearly as simple when I try to determine what $\sum^\infty_{k=1}{{1}\over{k^2+1}}$ equals.

Does there exist an elementary proof? or must I resort to using integrals?

I can show that $\sum^\infty_{k=1}{{1}\over{k^2}} = {{\pi^2}\over{6}}$ without to much hassle just using two representations of ${{sin(x)}\over{x}}$ but I cant find anything nearly as simple when I try to determine what $\sum^\infty_{k=1}{{1}\over{k^2+1}}$ equals.

Does there exist an elementary proof? or must I resort to using integrals?

I can show that $\sum^\infty_{k=1}{{1}\over{k^2}} = {{\pi^2}\over{6}}$ without to much hassle just using two representations of ${{sin(x)}\over{x}}$ but I cant find any proof nearly as simple when I try to determine what $\sum^\infty_{k=1}{{1}\over{k^2+1}}$ equals.

Does there exist an elementary proof? or must I resort to using integrals?

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Kate
  • 65
  • 1
  • 5

algebraic proof of an infinite sum

I can show that $\sum^\infty_{k=1}{{1}\over{k^2}} = {{\pi^2}\over{6}}$ without to much hassle just using two representations of ${{sin(x)}\over{x}}$ but I cant find anything nearly as simple when I try to determine what $\sum^\infty_{k=1}{{1}\over{k^2+1}}$ equals.

Does there exist an elementary proof? or must I resort to using integrals?