7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ What did you get for the ratio ? $\endgroup$ – Claude Leibovici May 14 '15 at 10:05
  • $\begingroup$ It converges for $q>p+1$, by comparison with the harmonic series. $\endgroup$ – Lucian May 14 '15 at 10:09
  • $\begingroup$ @Lucian Could you show me how did you compare it? $\endgroup$ – mobzopi May 14 '15 at 10:18
  • $\begingroup$ Hint: $(a+1)(a+2)(a+3)\cdots(a+n)=\dfrac{(n+a)!}{a!}$ $\endgroup$ – Lucian May 14 '15 at 10:23
  • $\begingroup$ Even if a is not an integer? $\endgroup$ – mobzopi May 14 '15 at 10:33
5
$\begingroup$

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 $q-p>1$.

$\endgroup$
  • $\begingroup$ (+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. $\endgroup$ – Jack D'Aurizio May 14 '15 at 14:03
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ (+1) Beat me by seconds! However, Gautschi's Inequalty might be useful, too. $\endgroup$ – robjohn May 14 '15 at 11:03
  • $\begingroup$ @robjohn: you are clearly right, I am adding your link to my answer to make it more complete. Do you mind? $\endgroup$ – Jack D'Aurizio May 14 '15 at 11:06
  • 1
    $\begingroup$ Of course not. Explanation is the purpose, and citation is the validation, of the answers here. $\endgroup$ – robjohn May 14 '15 at 11:27
  • $\begingroup$ What if $p=-1$ and $q=0$? $\endgroup$ – Alex M. May 14 '15 at 12:32
  • 1
    $\begingroup$ My mistake: $p>0$ by hypothesis. Otherwise, the series would have been convergent for integer values of $p<0$. $\endgroup$ – Alex M. May 14 '15 at 13:59
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Alex M. May 14 '15 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.