# $\sum_1^{\infty} \frac{(p+1)(p+2)(p+3)…(p+n)}{(q+1)(q+2)(q+3)…(q+n)}$ convergence

I need to determine for which values of $p$ and $q$, both greater than $0$, the following series converges: $$\sum_1^{\infty} \frac{(p+1)(p+2)(p+3)...(p+n)}{(q+1)(q+2)(q+3)...(q+n)}$$

I've tried using the ratio test, comparison test, and I've also tried partial fraction decomposition but I can't get to anything.

Could you give me a hint on how to solve this?

Any help will be appreciated

• What did you get for the ratio ? – Claude Leibovici May 14 '15 at 10:05
• It converges for $q>p+1$, by comparison with the harmonic series. – Lucian May 14 '15 at 10:09
• @Lucian Could you show me how did you compare it? – mobzopi May 14 '15 at 10:18
• Hint: $(a+1)(a+2)(a+3)\cdots(a+n)=\dfrac{(n+a)!}{a!}$ – Lucian May 14 '15 at 10:23
• Even if a is not an integer? – mobzopi May 14 '15 at 10:33

## 3 Answers

As you have probably noticed, the ratio test is useless here. The next thing to try when the ratio test fails is the Raabe-Duhame test. In our case this amounts to computing the following limit: $\lim \limits _{n \to \infty} n(\frac {q+n+1} {p+n+1} -1) = \lim \limits _{n \to \infty} \frac {(q-p)n} {p+n+1} = q-p$. Now, if $q-p < 1$ the series diverges and if $q-p>1$ the series converges. If $q-p=1$ then the Raabe-Duhamel test fails, but in this case the series becomes $\sum \frac {p+1} {p+1+n}$. Since $p \geq 0$, the series has the same behaviour as the series $\sum \frac 1 n$ which is divergent.

Therefore the series converges only if $q-p>1$.

• (+1) It is also interesting to compute the explicit value of the series under the assumption $q>p+1>0$. I did it through the Beta function, but maybe we can achieve the same by partial fraction decomposition and telescopic sums. – Jack D'Aurizio May 14 '15 at 14:03

We have: $$S(p,q)=\sum_{n\geq 1}\frac{(p+n)\cdot\ldots\cdot(p+1)}{(q+n)\cdot\ldots\cdot(q+1)}=\sum_{n\geq 1}\frac{\Gamma(p+n+1)\Gamma(q+1)}{\Gamma(q+n+1)\Gamma(p+1)}\tag{1}$$ but in virtue of the Euler product for the $\Gamma$ function (or by its log-convexity encoded in the Gautschi's inequality) we have: $$\frac{\Gamma(z+\alpha)}{\Gamma(z)}=\Theta(z^\alpha)\tag{2}$$ as $z\to +\infty$, hence the series is converging for $q>p+1>0$ by the p-test.

Moreover, in such a case: $$\begin{eqnarray*}S(p,q)&=&\frac{\Gamma(q+1)}{\Gamma(p+1)\Gamma(q-p)}\sum_{n\geq 1}B(p+n+1,q-p)\\&=&\frac{1}{B(p+1,q-p)}\int_{0}^{1}\sum_{n\geq 1}x^{p+n}(1-x)^{q-p-1}\,dx\\&=&\frac{1}{B(p+1,q-p)}\int_{0}^{1}x^{p+1}(1-x)^{q-p-2}\,dx\\[1em]&=&\frac{B(p+2,q-p-1)}{B(p+1,q-p)}=\color{red}{\frac{(p+1)}{q-(p+1)}}.\tag{3}\end{eqnarray*}$$

• (+1) Beat me by seconds! However, Gautschi's Inequalty might be useful, too. – robjohn May 14 '15 at 11:03
• @robjohn: you are clearly right, I am adding your link to my answer to make it more complete. Do you mind? – Jack D'Aurizio May 14 '15 at 11:06
• Of course not. Explanation is the purpose, and citation is the validation, of the answers here. – robjohn May 14 '15 at 11:27
• What if $p=-1$ and $q=0$? – Alex M. May 14 '15 at 12:32
• My mistake: $p>0$ by hypothesis. Otherwise, the series would have been convergent for integer values of $p<0$. – Alex M. May 14 '15 at 13:59

I think lucian is right.

Rewrite all as follows:

$\sum_{n=1}^\infty \frac{ (p+n)!}{p!} \frac{q!}{(q+n)!}$ and let $q=p+k$ the fact p has to be less than q is obvious.

$\sum_{n=1}^\infty \frac{ (p+n)!}{p!} \frac{q!}{(q+n)!} = \sum_{n=1}^\infty \prod _{j=1}^k \biggl(\frac {p+j}{p+n+j} \biggr) \leq (p+k)^k \sum_{n=1}^\infty \prod_{j=1}^k \frac{ 1}{p+n+j}$ $\leq (p+k)^k \sum_{n=1}^\infty \frac{1}{(p+n+1)^k}$ and now you have your classical $1/n^i$, and the answer is the one given by lucian.

• Please note that $p$ and $q$ are not necessarily natural numbers, therefore it might not make sense to speak about their factorials. A proof involving the $\Gamma$ function has been given above by @Jack D'Aurizio. – Alex M. May 14 '15 at 15:07