Is it true that $x\geq \sum_{k=1}^\infty k \prod_{n=1}^k \frac{x}{x+n}$ for $0<x<1$

Question

I would like to prove or disprove $x\geq \sum_{k=1}^\infty k \prod_{n=1}^k \frac{x}{x+n}$. From numerical examples it looks like the inequality holds. However, I have no idea how to prove it.

I have found a similar question sum of an infinite series $\sum_{k=1}^\infty \left( \prod_{m=1}^k\frac{1}{1+m\gamma}\right) $, where the function looks very similar to mine, but not the same.

Looking forward to any hints. Thanks in advance!

Answer 1

As you noticed, the equality hold and it hold for any $x >0$.

$$\prod_{n=1}^k\frac{x}{x+n}=\frac{x^k}{(x+1)_k}$$ where appear Pochhammer symbols and $$\sum_{k=1}^\infty k \prod_{n=1}^k \frac{x}{x+n}=\sum_{k=1}^\infty\frac{k\,x^k}{(x+1)_k}=\frac{x^2\,\Gamma (x)}{\Gamma (x+1)}=x$$

Answer 2

For a proof of $\sum_{k\geq 1}\frac{k x^k}{(x+1)_k}=x$ avoiding Complex Analysis, one might invoke Euler's Beta function.

$$\frac{1}{(x+1)_k}=\frac{\Gamma(x+1)}{\Gamma(x+k+1)}=\frac{1}{\Gamma(k)}B(k,x+1)=\frac{1}{(k-1)!}\int_{0}^{1}u^{k-1}(1-u)^x\,du.$$ Multiplying both sides by $kx^k$ and summing over $k\geq 1$ we get

$$ \sum_{k\geq 1}\frac{k x^k}{(x+1)_k} = \int_{0}^{1}(1-u)^x \sum_{k\geq 1}\frac{k x^k u^k}{(k-1)!}\cdot\frac{du}{u}=x\int_{0}^{1}(1-u)^x (1+ux)e^{ux}\,du$$ where the last integral does not really depend on $x$: it equals $1$ for any $x>0$, since the integrand function is $-\frac{d}{du}\left[(1-u)^{x+1}e^{ux}\right]$.