If $p > 0$, show that the series $\sum_{n=1}^{\infty} \frac {n!}{p(p+1) \cdots (p+n-1)}$ converges for $p > 2$ and diverges for $p \le 2$

Question

If $ p \in \Bbb Z$, $p > 0$, show that the following series converges for $p > 2$ and diverges for $p \le 2$ $$\sum_{n=1}^{\infty} \frac {n!}{p(p+1) \cdots (p+n-1)}$$

I'm having a lot of difficulty with this problem... I'm trying to simplify the denominator somehow so it's easier for me to visualize and solve but it's not going so well. Any tips on how to solve this?

Related: https://math.stackexchange.com/questions/821376/for-b-gt-2-verify-that-sum-n-1-infty-fracnbb1-bn-1-fra. — Martin R, Apr 02 '18 at 17:30
Is $p$ meant to be an integer, or arbitrary? There's a relatively straightforward argument in the integer case. (In fact, the divergence for $p=2$ can easily be used to show that half of the statement by comparison, and convergence for $p=3$ along with a similar comparison means you just have to look at $2\lt p\lt 3$ (if you're in the non-integer case). — Steven Stadnicki, Apr 02 '18 at 17:43
@StevenStadnicki it doesn't explicitly say, but I am assuming $p \in \Bbb Z$ — c87, Apr 02 '18 at 17:47
Possible duplicate of For $b \gt 2$ , verify that $\sum_{n=1}^{\infty}\frac{n!}{b(b+1)...(b+n-1)}=\frac{1}{b-2}$. — Nosrati, Dec 08 '18 at 13:33

score 1 · Answer 1 · answered Apr 02 '18 at 17:58

1

Big hint: since $p$ is 'constant', note that you can multiply by $\frac{(p-1)!}{(p-1)!}$, pull the numerator out of the sum, and get your series as $(p-1)!\sum_{n=1}^\infty\frac{n!}{(p+n-1)!}$. Now, this should simplify a lot...

answered Apr 02 '18 at 17:58

Steven Stadnicki

53,147

score 0 · Answer 2 · answered Apr 02 '18 at 18:30

For $p=2$, we have

$$\frac {n!}{p(p+1) ... (p+n-1)} = \frac{1}{n+1}$$

and hence the series diverges like the harmonic series.
For $0<p<2$, we can do a comparison with $p=2$ to conclude that the series also diverges.

Now, suppose $p>2$ and write $p=2+\epsilon$. We can write the term as

\begin{align}a_n=\frac {1}{ \frac{p}2\, \frac{p+1}3\, \dots \frac{p+n-2}n\, (p+n-1) } &= \frac {1}{ \left(1+\frac{\epsilon}2\right) \left(1+\frac{\epsilon}3\right) \dots \left(1+\frac{\epsilon}n\right)\, (p+n-1) } \\&= \frac{\frac{1}{n+1}}{ \left(1+\frac{\epsilon}2\right) \left(1+\frac{\epsilon}3\right) \dots \left(1+\frac{\epsilon}n\right)\, \left(1+\frac{\epsilon}{n+1}\right)\, }. \end{align}

Let $b_n = n\left(\frac{a_n}{a_{n+1}}-1\right)$. We can check that

\begin{align} b_n &= \frac{n}{n+1}\, (1+\epsilon). \end{align}

Then $b_n\to1+\epsilon >1$ as $n\to\infty$ and hence $\sum a_n$ converges by Raabe's test.

If $p > 0$, show that the series $\sum_{n=1}^{\infty} \frac {n!}{p(p+1) \cdots (p+n-1)}$ converges for $p > 2$ and diverges for $p \le 2$

2 Answers2