Interval of convergence of $\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!}$

Question

Given the series

$$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!} \quad \quad k \geq 1 $$ Find the interval of convergence.

I started by applying the Ratio test

$$ \lim_{n\to \infty}\left|\frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)(k+n)x^{n+1}}{(n+1)!}\cdot \frac{n!}{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}\right|$$

$$\lim_{n\to \infty}\left|\frac{(k+n)x}{(n+1)}\right|$$

to show that the series converges when $|x| \lt 1$.

However, when I test the end points of $(-1,1)$ for convergence, I end up with two series whose convergence I am unable to show. Namely, $$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)}{n!} $$

and $$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)(-1)^n}{n!} $$

How can I show that these two series converge or diverge?

Answer 1

Note that for $k\ge1$, we have $$ \frac{k(k+1)\cdots(k+n-1)}{n!}=\frac k1\frac{k+1}2\cdots\frac{k+n-1}{n}\ge1 $$ Thus, for $|x|=1$, the terms do not go to $0$.

Answer 2

Your series is $$ \sum_{n=0}^\infty\binom{-k}{n}(-x)^n=(1-x)^{-k} $$ This alone should show that there is no convergence at $x=1$ for positive $k$.

For the series at $x=-1$ consider that $$ \binom{-k}{n}(-1)^n=\binom{n+k-1}{n}=\binom{n+k-1}{k-1}=\frac{(n+1)(n+2)···(n+k-1)}{(k-1)!} $$ is a polynomial in $n$ of degree $k-1$.