For the past week, I've been mulling over this Math.SE question. The question was just to prove convergence of $$\sum\limits_{n=1}^\infty\frac{n!}{\left(1+\sqrt{2}\right)\left(2+\sqrt{2}\right)\cdots\left(n+\sqrt{2}\right)}$$ but amazingly Mathematica told me it had a remarkably simple closed form: just $1+\sqrt{2}$. After some fiddling, I conjectured for $a>1$:


I had been quite stuck until today when I saw David H's helpful answer to a similar problem. I have included a solution using the same idea, but I would be interested to know if anyone has another method.


Here is a completely elementary proof, which only needs introductory calculus concepts:

$$a_{n+1} = \frac{(n+1)!}{(a+1)(a+2)\dots(a+n+1)} = \frac{1}{a-1}\left(\frac{(n+1)!}{(a+1)(a+2) \dots (a+n)} -\frac{(n+2)!}{(a+1)(a+2) \dots (a+n+1)}\right) = b_{n+1} - b_{n+2}$$

where $a_n$ is the term of our series and $$b_n = \frac{1}{a-1}\left(\frac{n!}{(a+1)(a+2) \dots (a+n-1)}\right)$$ with $b_1 = \frac{1}{a-1}$

Thus the sum we seek, telescopes!

Giving us

$$\sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (b_{k} - b_{k+1}) = b_1 - b_{n+1}$$

Thus we just need to compute $\lim b_n$

(we basically need to show that $b_n \to 0$ to match the limit being $\frac{1}{a-1}$)

Now we have that $a \gt 1$, so let $a = 1 + x$ for $x \gt 0$.

$$ (a-1)b_n =\frac{n!}{(a+1)(a+2) \dots (a+n-1)} = \frac{1}{(1 + x/2)(1+x/3)\dots(1+x/n)} $$

Let $ M = \lceil x \rceil$ and consider the product

$$p_n = \prod_{k=M}^{n} \left(1 + \frac{x}{k}\right)$$

It is enough to show that $\log p_n \to \infty$ (that proves that $b_n \to 0$).

Now we have that $\dfrac{1}{1+t} \gt 1-t$ for $0 \lt t \le 1$ and thus integrating between $0$ and $y$ (where $y \le 1$) we get that

$$ \log(1+y) \ge y - \dfrac{y^2}{2}$$

Now $$\log p_n = \sum_{k=M}^{n} \log (1 + \frac{x}{k})$$

$$ \ge \sum_{k=M}^{n} (\frac{x}{k} - \frac{x^2}{2k^2})$$

This goes to $\infty$ as the harmonic series diverges, and the sum of reciprocals of the squares converges.

Thus the sum of your sequence is $$b_1 = \frac{1}{a-1}$$

  • 1
    $\begingroup$ An elementary path. +1 $\endgroup$ – Olivier Oloa Jan 14 '15 at 21:42
  • $\begingroup$ Very nice solution ! $\endgroup$ – S.Sundara Narasimhan Oct 1 '19 at 18:00
  • $\begingroup$ @S.SundaraNarasimhan: Thanks! $\endgroup$ – Aryabhata Oct 4 '19 at 1:23

The idea of this solution is to appeal to the Beta function and then to exchange the order of integration and summation (made possible by Fubini's theorem).

$$\begin{align} \sum\limits_{n=1}^\infty\prod\limits_{k=1}^n\frac{k}{k+a}&=\sum\limits_{n=1}^\infty\frac{n!}{(1+a)(2+a)\cdots(n+a)} \\&=\sum\limits_{n=1}^\infty\frac{\Gamma(n+1)\Gamma(1+a)}{\Gamma(n+a+1)} \\&=\Gamma(1+a)\sum\limits_{n=1}^\infty\frac{\Gamma(n+1)}{\Gamma(n+a+1)} \\&=\frac{\Gamma(1+a)}{\Gamma(a)}\sum\limits_{n=1}^\infty\frac{\Gamma(n+1)\Gamma(a)}{\Gamma(n+a+1)} \\&=a\sum\limits_{n=1}^\infty \operatorname{B}(n+1,a) \\&=a\sum\limits_{n=1}^\infty\int\limits_0^1t^n(1-t)^{a-1}\,dt \\&=a\int\limits_0^1\sum\limits_{n=1}^\infty t^n(1-t)^{a-1}\,dt \\&=a\int\limits_0^1\frac{t(1-t)^{a-1}}{1-t}\,dt \\&=a\int\limits_0^1t(1-t)^{a-2}\,dt \\&=a\operatorname{B}(2,a-1) \\&=\frac{a}{a(a-1)} \\&=\frac{1}{a-1} \end{align}$$

Note that we used $a>1$ when evaluating $\operatorname{B}(2,a-1)$, since the beta function is only defined when both arguments are greater than $0$.

A final note: the restriction $a>1$ is sharp in the sense that when $a=1$ the inside product simplifies to $$\prod\limits_{k=1}^n\frac{k}{k+1}=\frac{1}{n+1}$$ so the sum becomes $$\sum\limits_{n=1}^\infty\prod\limits_{k=1}^n\frac{k}{k+1}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots=\infty$$ and for $a<1$ in essence we can use the comparison test with the $a=1$ case to show that the sum diverges.

  • $\begingroup$ However, this is essentially the same answer given by David H to the linked question, and I bet the OP was looking for something different. $\endgroup$ – Jack D'Aurizio Jan 11 '15 at 20:50
  • $\begingroup$ @JackD'Aurizio Firstly, thank you for your answer. Secondly, I'm sorry for any confusion - I am both the author of this answer (motivated by David H's answer on the other question) and the OP of this question. $\endgroup$ – Peter Woolfitt Jan 11 '15 at 20:58
  • $\begingroup$ Oh, so sorry, I did not notice it :D $\endgroup$ – Jack D'Aurizio Jan 11 '15 at 21:00

Another possible trick is the following one. For any $a$ such that $\Re(a)>1$, let: $$ f(a) = \sum_{n=1}^{+\infty}\prod_{k=1}^{n}\frac{k}{k+a} $$ Since: $$\begin{eqnarray*} \frac{1}{a+1}\, f(a+1) &=& \sum_{n=1}^{+\infty}\frac{n!}{(a+1)(a+2)\cdot\ldots\cdot(a+n+1)}\\&=&\sum_{n=1}^{+\infty}\frac{(n-1)!}{(a+2)\cdot\ldots\cdot(a+n)}\left(\frac{1}{a+1}-\frac{1}{a+n+1}\right)\end{eqnarray*}$$ it follows that: $$\frac{1}{a+1}\,f(a+1) = f(a+1)-f(a+2)\tag{1} $$ or: $$ a\, f(a+1) = (a+1)\, f(a+2)\tag{2} $$ So, given that $ z\,f(z+1) $ is an analytic function on $\Re(z)>1$, we have that $z\,f(z+1)$ is constant, and: $$ 2\, f(3) = \sum_{n=1}^{+\infty}\frac{12}{(n+1)(n+2)(n+3)}=\sum_{n=1}^{+\infty}\left(\frac{6}{n+1}-\frac{12}{n+2}+\frac{6}{n+3}\right)=\color{red}{1}\tag{3} $$ by the telescopic property. This gives that over $\Re(a)>1$, $$ f(a) = \color{red}{\frac{1}{a-1}}\tag{4}$$ holds.

  • 2
    $\begingroup$ Thanks! And that's a super sweet factorization. $\endgroup$ – Peter Woolfitt Jan 11 '15 at 22:44
  • 2
    $\begingroup$ Maybe I'm stupid, but how does it follow that $z\,f(z+1)$ is constant? What equation (2) says is that it is periodic with period $1$, but there are many non-constant analytic functions with period $1$... $\endgroup$ – Hans Lundmark Jan 12 '15 at 6:28
  • $\begingroup$ @HansLundmark: Right! Why is it constant Jack? -1 till this is resolved. $\endgroup$ – Aryabhata Jan 12 '15 at 17:45
  • $\begingroup$ @HansLundmark: you are right, but the point here is that with a minor tweak, we can prove that $g(z)=z f(z+1)$ satisfies the functional equation: $$ g(a) = g(a+1/m) $$ for any $m\in\mathbb{N}^*$, so, by continuity and density, $g$ is constant. $\endgroup$ – Jack D'Aurizio Jan 12 '15 at 19:48

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.