Analytically Solving Complex Expressions with Wolfram Alpha

[Office_Shredder]

I'm noticing wolfram alpha has the amazing ability to analytically solve
\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}

Anyone know how to do this, and if it's also possible to deal with higher order guys (like it also can do 1/(n⁴+a²), but it's a way more complicated expression to the point where I'm staring at it wondering if it's actually a real number)

 

[mathman]

It is a real number. However I have no idea how to calculate it.

 

[Mandelbroth]

I'm noticing wolfram alpha has the amazing ability to analytically solve
\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}

Anyone know how to do this, and if it's also possible to deal with higher order guys (like it also can do 1/(n⁴+a²), but it's a way more complicated expression to the point where I'm staring at it wondering if it's actually a real number)

I'd split it into partial sums and then evaluate.

Hint:
$\frac{1}{n^2+a^2}=\frac{1}{(n+ai)(n-ai)}$, where i is the imaginary unit.

Higher order expressions will undoubtedly be a lot more complicated. Though, given n and a are real numbers, the sum will also be real, so they won't necessarily be more complex (heh. See what I did there? ).

 

[Office_Shredder]

Higher order expressions will undoubtedly be a lot more complicated. Though, given n and a are real numbers, the sum will also be real, so they won't necessarily be more complex (heh. See what I did there? ).

If you split it into two fractions of degree one neither of your series converge anymore.

By "wondering if it's a real number" I meant "gee wolfram alpha has a lot of 4th roots of -1 in that expression" not "I literally don't know if it's a real number"

 

[dextercioby]

This formula right here explains where the Wolframalpha result comes from

http://functions.wolfram.com/ElementaryFunctions/Coth/06/05/0001/

All you have to do now is to series expand the Coth function to get the RHS.

 

[Office_Shredder]

This formula right here explains where the Wolframalpha result comes from

http://functions.wolfram.com/ElementaryFunctions/Coth/06/05/0001/

All you have to do now is to series expand the Coth function to get the RHS.

Oh wow that was easy. I was too busy approaching the problem from the wrong side of the equation. Thanks

 

[dextercioby]

It is actually not easy, not to me. I can't find a proof for the series expansion. :D