# $\sum_{n=1} ^{\infty} (-1)^{n-1} \frac{(1+p)(2+p)\cdots(n+p)}{n! n^q} \text{converges}\impliedby p=?,q=?$

Could you help me with this?

I need to check for what $p, q \in \mathbb{Z}$ does the series $\sum_{n=1} ^{\infty} (-1)^{n-1} \frac{(1+p)(2+p)\cdots(n+p)}{n! n^q}$ converge (conditionally or absolutely).

I don't know what criterion to use. I've tried Cauchy's, D'Alembert, Raabe's and each time the result is dependent on $n$.

I've found something like this:

If $(a_n)$ is a number sequence and if $\sum _{n = 1} ^{\infty} \frac{a_n}{n^x}$ is convergent for certain $x^*$, then it is convergent for any $x \ge x^*$. (It follows from Abel's criterion, because we multiply the terms by decreasing factors $\frac{1}{n^{x-x*}}$ )

I've been thinking about putting $a_n = \frac{(1+p)(2+p)\cdots(n+p)}{n!}$ but that doesn't make things easier, does it?

I'd really appreciate any hints.

$$(1)$$

It converges absolutely when $q> p+1$. Multiplying top and bottom by $p!:$

$$\sum_{n\geq 1}\frac{(n+1)\cdots (n+p)}{p!n^{q}}$$

The numerator can be written as $n^p+r(n)$ where $r$ is some polynomial with $\deg r<p:$

$$\sum_{n\geq 1}\frac{1}{p!}\left(n^{p-q}+\frac{r(n)}{n^q}\right)$$

By comparison with the Harmonic $p$-series this diverges when $q\leq p+1$ and converges otherwise.

$$(2)$$

It converges conditionally when $q\geq p+1$. This is easily seen by using what we have previously done:

$$\sum_{n\geq 1}\frac{(-1)^{n-1}}{p!}\left(n^{p-q}+\frac{r(n)}{n^q}\right)$$

will converge if $q=p+1$ since the alternating harmonic series converges. When $q\leq p$ on the other hand, $n^{p-q}\geq 1$ so we cannot have convergence.

First, show that $${(1+p)(2+p)\times\cdots\times(n+p)\over n!}$$ is comparable to $n^p$ for large values of $n$. Then all you have to do is decide on the convergence of $\sum(-1)^nn^{p-q}$

If you tried Stirling approximation:

$n!\sim \sqrt{2\pi n}(\dfrac{n}{e})^n$, then

$\dfrac{(n+p)!}{n!p!}n^{-q}\sim\dfrac{1}{\exp(p)}\sqrt{\dfrac{n+p}{n}}\dfrac{(n+p)^{n+p}}{n^{n+q}p!} = \exp(-p)(1+\dfrac{p}{n})^{n+q+1/2}\dfrac{1}{p!}n^{p-q}\sim \dfrac{1}{p!}n^{p-q}$.

And $\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{p!}n^{p-q}$ will absolutely converge if $q\ge p+2$, and will conditionally converge if $q = p+1$.