Prove that $f(x)=\sum_{i=0}^{M-1}(-1)^i\binom{M}{i+1}\frac{1}{1+ix\Delta}$ increases with $x$?

Question

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 $x$?

Answer 1

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.

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Proof of the 1^st 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 2^nd 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*}

Answer 2

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.