<s>[INST] Interval of convergence of $\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!}$ 
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?
 [/INST] 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 $k-1$.
</s>