Proving an inequality on $ \int_0^\infty \frac{e^{-tx}}{1+e^{-t}}dt$

Question

EDIT : I have posted a proof below that needs to be reviewed.

Some definitions

Let $$\begin{array}{ccccc} f & : & \mathbb R_+^* & \to & \mathbb R_+^* \\ & & x & \mapsto & \int_0^\infty \frac{e^{-tx}}{1+e^{-t}}dt \\ \end{array}$$

Let $$\begin{array}{ccccc}\forall n\in \mathbb N, K_n & : & \mathbb R_+^* & \to & \mathbb R_+^* \\ & & x & \mapsto & \frac{\Gamma(n+1) \Gamma(x)}{\Gamma(n+x+1)} \\ \end{array} $$

Note that equivalently $$ \forall n, \forall x>0, K_n(x)=\frac{n!}{x\times(x+1)\times\ldots\times(x+n)}$$

Question

Prove that $$ \forall n \in \mathbb N, \forall x>0,f(x)\leq \frac{K_{n+1}}{2^{n+1}}+\sum_{k=0}^{n} \frac{K_k(x)}{2^{k+1}}$$

Context

I've been working thoroughly on $f$ over the past few days (see my other posts) in the scope of an assignment I got.

Here are some properties I have shown so far:

$$\forall x>0, f(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{n+x}$$

$$\forall x>0, f(x)= \sum_{k=0}^{\infty} \frac{K_k(x)}{2^{k+1}}$$

What I've tried

Given the last identity, my question boils down to proving that $$ \forall n, \forall x>0 \;, \; 0 \leq \frac{K_{n+1}}{2^{n+1}} - \sum_{k=n+1}^{\infty} \frac{K_k(x)}{2^{k+1}} $$

I don't even know how to start... Induction maybe (since it's valid $\forall n$). I failed proving the case $n=0$ as well.

Thanks for any suggestion.

Answer 1

I think I got the proof, in an (overly) simple way that makes me question its validity.

Let $n\in \mathbb N$ and $x>0$

We have that $$f(x)= \sum_{k=0}^{n} \frac{K_k(x)}{2^{k+1}}+\lim\limits_{M \to \infty}\sum_{k=n+1}^{M} \frac{K_k(x)}{2^{k+1}}$$

For some fixed $M>n+1$ we have that $$\sum_{k=n+1}^{M} \frac{K_k(x)}{2^{k+1}}=\sum_{k=n+1}^{M}\frac{1}{2^{k+1}}\times\frac{k\times(k-1)\ldots\times(n+2)}{(x+k)\times(x+k-1)\ldots\times(x+n+2)}\times K_{n+1}(x)$$

Since $x>0$, $$\sum_{k=n+1}^{M}\frac{1}{2^{k+1}}\times\frac{k\times(k-1)\ldots\times(n+2)}{(x+k)\times(x+k-1)\ldots\times(x+n+2)}\times K_{n+1}(x) \leq \frac{K_{n+1}(x)}{2^{n+1}}\times \sum_{k=1}^{\infty}\frac{1}{2^k}$$

Hence $$\sum_{k=n+1}^{M} \frac{K_k(x)}{2^{k+1}} \leq \frac{K_{n+1}(x)}{2^{n+1}}$$

This identity holding for all $M$, taking the limit on $M$ yields: $$\sum_{k=n+1}^{\infty} \frac{K_k(x)}{2^{k+1}} \leq \frac{K_{n+1}(x)}{2^{n+1}}$$

Hence $$f(x) \leq \sum_{k=0}^{n} \frac{K_k(x)}{2^{k+1}} + \frac{K_{n+1}(x)}{2^{n+1}}$$

Answer 2

This an approach which tends to the form of the expected inequality :