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?

  • 2
    $\begingroup$ Related: math.stackexchange.com/questions/821376/…. $\endgroup$
    – Martin R
    Apr 2 '18 at 17:30
  • $\begingroup$ 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). $\endgroup$ Apr 2 '18 at 17:43
  • $\begingroup$ @StevenStadnicki it doesn't explicitly say, but I am assuming $p \in \Bbb Z$ $\endgroup$
    – c87
    Apr 2 '18 at 17:47
  • $\begingroup$ Possible duplicate of For $b \gt 2$ , verify that $\sum_{n=1}^{\infty}\frac{n!}{b(b+1)...(b+n-1)}=\frac{1}{b-2}$. $\endgroup$
    – Nosrati
    Dec 8 '18 at 13:33

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...


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.


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