Convergence of a (barely converging?)series

Question

Is the following series convergent?

$$ \sum_{n\geq N}\left(1-\frac{a}{N}\right)\left(1-\frac{a}{N+1}\right)\cdots\left(1-\frac{a}{n}\right) $$

The assumptions are $N>a>1$ and a is real.

I suspect it barely does, but I cannot get a good start on it.

Answer 1

The given series can be written as

$$\begin{eqnarray*} \sum_{n\geq N}\frac{\Gamma(n+1-a)\,\Gamma(N)}{\Gamma(N-a)\,\Gamma(n+1)}&=&\frac{\Gamma(N)}{\Gamma(N-a)}\sum_{n\geq N}\frac{\Gamma(n+1-a)}{\Gamma(n+1)}\\&=&\frac{\Gamma(N)}{\Gamma(N-a)\Gamma(a)}\sum_{n\geq N}B(n+1-a,a)\\&=&\frac{\Gamma(N)}{\Gamma(N-a)\Gamma(a)}\int_{0}^{1}\sum_{n\geq N}(1-x)^{a-1}x^{n-a}\,dx\\&=&\frac{\Gamma(N)}{\Gamma(N-a)\Gamma(a)}\int_{0}^{1}x^{N-a}(1-x)^{a-2}\,dx\\&=&\frac{\Gamma(N)}{\Gamma(N-a)\Gamma(a)}B(N+1-a,a-1)\\&=&\color{red}{\frac{N-a}{a-1}}\end{eqnarray*}$$

by exploiting Euler's Beta function. The given series is convergent since $\frac{\Gamma(n+1-a)}{\Gamma(n+1)}$ behaves like $\frac{1}{n^a}$ (that is summable for $a>1$) by Gautschi's inequality.

Answer 2

Not only does it converge, the sum can be evaluated in closed form: $$ \frac{N-a}{a-1}$$

EDIT: In fact, it telescopes: the partial sum

$$ \sum_{n=N}^M \frac{\Gamma(N) \Gamma(n+1-a)}{\Gamma(N-a)\Gamma(n+1)} = \frac{N-a}{a-1} - \frac{\Gamma(N) \Gamma(M+2-a)}{(a-1)\Gamma(M+1)\Gamma(N-a)}$$