Sum: $\sum_{n=1}^\infty\prod_{k=1}^n\frac{k}{k+a}=\frac{1}{a-1}$

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 $a>1$:

$$\sum\limits_{n=1}^\infty\prod\limits_{k=1}^n\frac{k}{k+a}=\sum\limits_{n=1}^\infty\frac{n!}{(1+a)(2+a)\cdots(n+a)}=\frac{1}{a-1}$$

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.

Solution 1:

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.

Solution 2:

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 $$b_1 = \frac{1}{a-1}$$

Solution 3:

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.