Prove this summation equivalence as an identity?

Question

  ∞
  ∑ [b] / [n² − b² ] =
n=b+1

  2b
  ∑ 1 / [2n]
n=1

http://i20.photobucket.com/albums/b249/LindySoul/mathematics/Summation/091a156befacae4fdacb861ebcfc7b36.png

I derived this myself earlier today... I am uncertain if people know or acknowledge it

Answer 1

Actually, this one's pretty easy.

[n=b+1, ∞]∑b/(n²-b²)

We start by decomposing b/(n²-b²) as 1/2 (1/(n-b) - 1/(n+b)) (this was obtained through partial fraction decomposition). Then we have:

1/2 [n=b+1, ∞]∑(1/(n-b) - 1/(n+b))

Writing this as a limit:

1/2 [k→∞]lim [n=b+1, k]∑(1/(n-b) - 1/(n+b))

Breaking up the series into two series (note that we could not do this without explicitly writing the limit sign, since the two series individually do not converge, only their difference does):

1/2 [k→∞]lim ([n=b+1, k]∑1/(n-b) - [n=b+1, k]∑1/(n+b))

Re-indexing both the first and the last series:

1/2 [k→∞]lim ([n=1, k-b]∑1/n - [n=2b+1, k+b]∑1/n)

Breaking off the non-canceling parts of the two series:

1/2 [k→∞]lim ([n=1, 2b]∑1/n + [n=2b+1, k-b]∑1/n - [n=2b+1, k-b]∑1/n - [n=k-b+1, k+b]∑1/n)

Canceling:

1/2 [k→∞]lim ([n=1, 2b]∑1/n - [n=k-b+1, k+b]∑1/n)

Re-indexing the last series:

1/2 [k→∞]lim ([n=1, 2b]∑1/n - [n=1, 2b]∑1/(n+k-b))

Now evaluating the limit (the interchange of sums and limits is justified since only finitely many terms are being summed, and the limits of summation no longer depend on k):

1/2 ([n=1, 2b]∑1/n - [k→∞]lim [n=1, 2b]∑1/(n+k-b))
1/2 ([n=1, 2b]∑1/n - [n=1, 2b]∑[k→∞]lim 1/(n+k-b))
1/2 ([n=1, 2b]∑1/n - [n=1, 2b]∑0)
1/2 [n=1, 2b]∑1/n
[n=1, 2b]∑1/(2n)

Q.E.D.