# Inequality involving exponential partial sums

Consider the exponential partial sums $E_n(x) = \sum_{i=0}^n \frac{x^i}{i!}$. I want to prove that for all $x \ge 0$: $$2 \frac {E_{n-1}(x)} {E_n(x)} \ge \frac {E_{n}(x)} {E_{n+1}(x)} + \frac {E_{n-2}(x)} {E_{n-1}(x)}$$

My approach so far

First observe that $E_{n-1}(x) = E_{n}(x) - \frac {x^n}{n!}$. So the inequality becomes: $$2 \frac {E_{n}(x) - \frac {x^n}{n!}} {E_n(x)} \ge \frac {E_{n+1}(x) - \frac {x^{n+1}}{(n+1)!}} {E_{n+1}(x)} + \frac {E_{n-1}(x) - \frac {x^{n-1}}{(n-1)!}} {E_{n-1}(x)}$$ which leads to $$2 \frac {\frac {x^n}{n!}} {E_n(x)} \le \frac {\frac {x^{n+1}}{(n+1)!}} {E_{n+1}(x)} + \frac {\frac {x^{n-1}}{(n-1)!}} {E_{n-1}(x)}$$

So all we need to show is that $\frac {x^n} {n! E_n(x)}$ is convex in $n$. Unfortunately, I didn't have much luck going forward. A good direction could be to use the fact that $n! E_n(x) = e^x \Gamma(n+1,x)$, where $\Gamma(n+1,x) = \int_x^\infty t^n e^{-t} \textrm{dt}$ is the incomplete gamma function. I feel that this way, I will be able to prove the inequality analytically without working painfully with factorials and large sums. So it suffices to show that the following is convex as a function of $n$: $$\frac {x^n} {\int_x^\infty t^n e^{-t} \textrm{dt}}$$

Any ideas on how to continue? Unfortunately, derivatives of the incomplete gamma function with respect to $n$ are not as nice as those with respect to $x$.

• Guess you want $x \ge 0$. Otherwise it doesn't hold anyway. – Macavity Mar 15 '15 at 6:16
• You are correct. I edited the description. – thelionkingrafiki Mar 15 '15 at 12:44
• A quick calculation using the derivative under the integral rule suggests to me, if $f(\nu):=x^\nu/\int_x^\infty t^\nu e^{-t}\text{d}t$, then $f''(\nu)\leq0$, meaning $f$ is in fact concave. I may have made an error, but it should be easy for you to check whether or not this is the case. – Stromael Mar 16 '15 at 14:01
• I think there is some error in your calculations. The first derivative is negative as $f$ is decreasing but the second derivative should be positive. Here is an example for $x=5$ to see that this is the case: wolframalpha.com/input/… – thelionkingrafiki Mar 16 '15 at 14:11

Following thelionkingrafiki's reduction, we only need to show that, for $x > 0$ and $n \ge 1$, $$\frac{a_{n-1}}{E_{n-1}} +\frac{a_{n+1}}{E_{n+1}} -2\frac{a_{n}}{E_n} > 0,$$ where $a_{n} = x^n/n!$ and $E_n = \sum_{k=0}^n a_k$.
It turns out the inverse $E_n/a_n$ is easier to handle, so we shall define $$y_n \equiv \frac{E_n}{a_n} - 1.$$ and it can be shown by direct expansion that our statement is equivalent to $$(1 + 2 \, y_{n-1} - y_n)(y_{n-1} + y_{n+1} - 2 \, y_n) < 2 \, (y_n - y_{n-1})^2.$$
Now, by expanding the sum we have \begin{align} y_n &= \frac{n}{x} + \frac{n(n-1)}{x^2} + \frac{n(n-1)(n-2)}{x^3} + \cdots + \frac{n!}{x^n} \\ y_n - y_{n-1} &= \frac{1}{x} + \frac{2\,(n-1)}{x^2} + \frac{3\,(n-1)(n-2)}{x^3} + \cdots + \frac{n!}{x^n} \\ &= \frac{1}{x}\left( 1 \, b_1 + 2 \, b_2 + 3 \, b_3 + \cdots + n \, b_n \right), \end{align} where, $b_1 = 1$, and $$b_k = \frac{ (n-1)\cdots (n - k + 1) } { x^{k+1} },$$ for $k \ge 2$.