I am trying to confirm a Quantum-Mechanics related calculation and have stumbled upon this infinite ,I think, sum :
$$K_0 = \sum_{n=0}^{\infty}n\frac{(2h+n-1)!}{n!(2h-1)!}\frac{\tanh^{2n}(at)}{\cosh^{4h}(at)}$$
I already know the answer to be $$K_0= 2h\sinh^2(at)$$
Initially, I thought I write the sum in the form of a binomial expansion after breaking the $\tanh$ term into $\frac{sinh}{cosh}$ and through a commonly used method (e.g. set $\sinh^2(at)$ equalt to a parameter, say $p$, drop the first term of the sum and then make a variable change $n = n'+1$, rearrange the ''n by k'' term to match the coefficients ) get my result. However,my final result is $$ K_0 = \frac{p}{1+p}\sum_{n=0}^{\infty}\frac{(2h+n)!}{n!(2h)!}\frac{p^n}{(1+p)^n} =\tanh^2(at)\sum_{n=0}^{\infty}\frac{(2h+n)!}{n!(2h)!}\frac{p^n}{(1+p)^n} $$ which is off the mark I am pretty sure. I also tried to use the exponential definitions but I got lost very quickly. Any help or hints would be much appreciated, thanks!