Inequality $(\sum_{k=0}^{2n-1}x^k / k!)(\sum_{k=0}^{2n-1}(-x)^k / k!)\leq1$ for all $x\in\mathbb R$

Question

I am trying to find a proof of the following inequality $$\left(\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}\right)\left(\sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}\right)\leq1$$ for all $x\in\mathbb R$ and for all $n\in\mathbb Z_+$.

Does anyone know a proof for this without using derivative: for example by induction?....

Answer 1

This can be proved purely combinatorially using nothing more than the binomial theorem. In fact, I claim that the polynomial $P_n(x)=\left(\sum\limits_{k=0}^{2n-1} \frac{x^k}{k!}\right)\left(\sum\limits_{\ell=0}^{2n-1}(-1)^\ell \frac{x^\ell}{\ell!}\right)-1$ is even, has no constant term, and has no positive coefficients, which is sufficient to show that it is never positive.

The proof of this fact mostly recapitulates the formal-power-series proof that $e^xe^{-x}=1$. The catch is that some of the limits on our sums are different, in a way that makes it messier.

To start with, we'll multiply the two sums together, let $m=k+\ell$, and rewrite the whole sum in terms of $m$ and $\ell$. We have to be careful about the limits of the sum when we do this; the square in $k$ and $\ell$ turns into a diamond in $m$ and $\ell$, meaning the sum breaks up into two pieces. We have \begin{align} P_n(x)&=\left(\sum\limits_{k=0}^{2n-1} \frac{x^k}{k!}\right)\left(\sum\limits_{\ell=0}^{2n-1}(-1)^\ell \frac{x^\ell}{\ell!}\right)-1\\ &=\sum_{k=0}^{2n-1}\sum_{\ell=0}^{2n-1} (-1)^\ell\frac{x^k}{k!}\frac{x^\ell}{\ell!}-1\\ &=\sum_{m=0}^{2n-1} \sum_{\ell=0}^m (-1)^\ell \frac{x^m}{\ell!(m-\ell)!}-1+\sum_{m=2n}^{4n-2}\sum_{\ell=m-2n+1}^{2n-1} (-1)^\ell\frac{x^m}{\ell!(m-\ell)!}\\ &=\underbrace{\sum_{m=0}^{2n-1}\frac{x^m}{m!} \sum_{\ell=0}^m (-1)^\ell \binom{m}{\ell}-1}_{(*)}+\underbrace{\sum_{m=2n}^{4n-2}\frac{x^m}{m!}\sum_{\ell=m-2n+1}^{2n-1}(-1)^\ell \binom{m}{\ell}}_{(**)} \end{align}

We'll consider the two braced pieces of this sum separately.

First, I claim that the sum $(*)$ is identically zero. To see this, note that the coefficient $\sum_{\ell=0}^m (-1)^\ell \binom{m}{\ell}$ of $\frac{x^m}{m!}$ is the binomial expansion of $(1-1)^m$, and so it vanishes except when $m=0$. But when $m=0$, the coefficient of $x^m$ is equal to $1$, so it cancels with the subtracted-off $1$, leaving the entire expression identically zero as desired.

Now, we'll examine $(**)$. Our goal is to show that the coefficient $\sum\limits_{\ell=m-2n+1}^{2n-1}(-1)^\ell \binom{m}{\ell}$ of $\frac{x^m}{m!}$ is zero whenever $m$ is odd, and negative whenever $m$ is even; this will complete the proof. Note that this coefficient represents the middle terms of the binomial expansion of $(1-1)^m$; since the entire expansion sums to zero, we can write it in terms of the outer terms, which are friendlier: $$ \sum_{\ell=m-2n+1}^{2n-1}(-1)^\ell \binom{m}{\ell}=-\sum_{\ell=0}^{m-2n} (-1)^\ell \binom{m}{\ell}-\sum_{\ell=2n}^m (-1)^\ell \binom{m}{\ell} $$ Now, we use the identity $\binom{m}{\ell}=\binom{m}{m-\ell}$ to combine these two sums. When $m$ is odd, the factor of $(-1)^\ell$ means they combine with opposite sign and therefore vanish.

When $m$ is even, they combine with the same sign, yielding $$ -2\sum_{\ell=0}^{m-2n} (-1)^\ell \binom{m}{\ell} $$ So we need only show that the sum $\sum_{\ell=0}^{m-2n} (-1)^\ell \binom{m}{\ell}$ is positive when $m$ is even and $2n \leq m \leq 4n-2$. But with these bounds on $m$, $m-2n < \frac{m}{2}$, meaning the terms of the sum increase in absolute value as $\ell$ increases. Since $m$ is even, the last term in the sum will be positive, meaning that we can pair each negative term in the sum off with a larger positive term. Thus the entire sum is positive, and we are done.

Answer 2

By Taylor series with Lagrange remainder $$ e^x=\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\frac{(e^x)^{(2n)}|_t}{2n!}x^{2n}=\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\frac{e^t}{2n!}x^{2n} $$ where $|t|<|x|$. And $$ e^{-x}=\sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}+\frac{(e^{-x})^{(2n)}|_s}{2n!}x^{2n}=\sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}+\frac{e^{-s}}{2n!}x^{2n} $$ where $|s|<|x|$. So we have $$ \sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}<e^x\quad\text{and}\quad \sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}<e^{-x} $$ for all $x\in\mathbb R$. Since $$ \sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}=2\sum\limits_{k=0}^{n-1}\frac{x^{2k}}{2k!}>0 $$ for all $x\in\mathbb R$, $\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}$ and $\sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}$ can't be both negative. If one of them is non positive, then $$ \left(\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}\right)\left(\sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}\right)\leqslant 0<1 $$ If both of them are positive, then $$ \left(\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}\right)\left(\sum\limits_{k=0}^{2n-1}\frac{(-x)^k}{k!}\right)\leqslant e^{x}e^{-x}=1 $$

Answer 3

Solution 1. We introduce two sequences of functions $T_n(x)$ and $I_n(x)$ by

$$ T_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!} \quad \text{and} \quad I_n(x) = \int_{x}^{\infty} \frac{t^n}{n!} e^{-t} \, dt. $$

Applying integration by parts, we obtain the following recurrence relation:

\begin{align*} I_n (x) &= \left[ -\frac{t^n}{n!}e^{-t} \right]_{t=x}^{t=\infty} + \int_{x}^{\infty} \frac{t^{n-1}}{(n-1)!}e^{-t} \, dt \\ &= \frac{x^n}{n!}e^{-x} + I_{n-1}(x). \end{align*}

Iterating this, we obtain $I_n (x) = T_n(x)e^{-x}$ and hence

$$ T_n(x) = e^x I_n (x). \tag{1} $$

Now assume that $n$ is odd. Then $ t \mapsto (t^n / n!)e^{-t}$ is positive when $t > 0$ and negative when $t < 0$. Thus we find that

1. $I_n (x)$ achieves the global maximum $I_n (0) = 1$ at $x = 0$,
2. $I_n (x) > 0$ when $x \geq 0$.

These imply that $0 < I_n(|x|) \leq 1$ and $I_n(-|x|) \leq 1$. Therefore

$$ T_n(x) T_n(-x) = I_n (|x|) I_n(-|x|) \leq 1. $$

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Solution 2. I guess this is not the kind of solution that OP wants to see, but actually we have a very straightforward proof if we use differentiation appropriately.

Let $C_n(x)$ and $S_n(x)$ by

$$ C_n(x) = \sum_{k=0}^{n} \frac{x^{2k}}{(2k)!} \quad \text{and} \quad S_n(x) = \sum_{k=0}^{n} \frac{x^{2k+1}}{(2k+1)!}. $$

Using the same notation $T_n(x)$ as in Solution 1, we find that

1. $T_{2n+1}(x) = C_n(x) + S_n(x)$ and $T_{2n+1}(-x) = C_n(x) - S_n(x)$. Consequently, $$ T_{2n+1}(x) T_{2n+1}(-x) = C_n(x)^2 - S_n(x)^2.$$
2. $ S_n(x)' = C_n(x)$ and $C_n(x)' = S_{n-1}(x)$.

Now WLOG we assume that $x > 0$. Then

\begin{align*} (T_{2n+1}(x)T_{2n+1}(-x))' &= (C_n(x)^2 - S_n(x)^2)' \\ &= -2C_n(x)\{ S_n(x) - S_{n-1}(x) \} \\ & < 0. \end{align*}

Therefore $T_{2n+1}(x)T_{2n+1}(-x)$ is strictly decreasing on $[0, \infty)$ and the claim follows. ////

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Remark. (1) Using the same argument, we can also show that $T_{2n}(x)T_{2n}(-x) \geq 1$.

(2) Using the intermediate result, we can also confirms @Micah's observation that $P_n(x)$ is an even polynomial with no positive coefficient: \begin{align*} P_{n}(x) &= T_{2n-1}(x)T_{2n-1}(-x) - 1 \\ &= \int_{0}^{x} \{ T_{2n-1}(t)T_{2n-1}(-t) \}' \, dt \\ &= - 2 \int_{0}^{x} \frac{t^{2n-1}}{(2n-1)!}C_{n-1}(t) \, dt \\ &= - \sum_{k=0}^{n-1} \frac{x^{2n+2k}}{(2n-1)!(2k)!(n+k)}. \end{align*}