Prove that $f(x)=\sum_{i=0}^{M-1}(-1)^i\binom{M}{i+1}\frac{1}{1+ix\Delta}$ increases with $x$? How to prove that $f(x)=\sum_{i=0}^{M-1}(-1)^i\binom{M}{i+1}\frac{1}{1+ix\Delta}$, where $\Delta>0,M>1$, increases  with   $x$?
 A: Let us define $\varphi$ by
$$ \varphi(u) = \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} u^i. $$
By a simple computation
$$ \varphi(u)
= \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} \frac{M}{i+1} u^i
= \int_{0}^{1} M(1 - su)^{M-1} \, ds, $$
we find that $\varphi(u)$ decreases on $[0, 1]$. And this function is related to our $f$ by
$$
f(x)
= \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} \frac{1}{1+ix\Delta}
= \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} \int_{0}^{1} t^{ix\Delta} \, dt
= \int_{0}^{1} \varphi(t^{x\Delta}) \, dt.
$$
Since $x \mapsto t^{x\Delta}$ decreases for each $t \in (0, 1)$, it follows that $x \mapsto \varphi(t^{x\Delta})$ increases. Therefore the claim follows.

Proof of the 1st equality. Notice that
$$ \binom{M}{i+1} = \binom{M-1}{i}\frac{M}{i+1} \quad \text{and} \quad \int_{0}^{1} s^{i} \, ds = \frac{1}{i+1}. $$
Then
\begin{align*}
\varphi(u)
&= \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} \frac{M}{i+1} u^i \\
&= \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} M u^i \int_{0}^{1} s^{i} \, ds \\
&= M \int_{0}^{1} \left( \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} u^i s^{i} \right) \, ds \\
&= M \int_{0}^{1} (1 - su)^{M-1} \, ds,
\end{align*}
where in the last line we utilized the binomial theorem to simply the sum.
Proof of the 2nd equality. Again, we know that
$$ \int_{0}^{1} t^{ix\Delta} \, dt = \frac{1}{1+ix\Delta}. $$
So we have
\begin{align*}
f(x)
&= \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} \frac{1}{1+ix\Delta} \\
&= \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} \int_{0}^{1} t^{ix\Delta} \, dt \\
&= \int_{0}^{1} \left( \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} t^{ix\Delta} \right) \, dt \\
&= \int_{0}^{1} \varphi(t^{x\Delta}) \, dt.
\end{align*}
A: I don't think that
it's true.
For
$m=3$
with $\delta = 1$,
$f(x)
=1-\dfrac{3}{1+x}+\dfrac{3}{1+2x}
=1-3\dfrac{1+2x-(1+x)}{(1+x)(1+2x)}
=1-3\dfrac{x}{(1+x)(1+2x)}
$.
If
$g(x)
=\dfrac{x}{(1+x)(1+2x)}
$,
then,
according to Wolfy,
$g'(x)
=\dfrac{1 - 2 x^2}{(x + 1)^2 (2 x + 1)^2}
$
and this changes sign at
$1/\sqrt{2}$
so it is not monotonic.
