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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 $x$.

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  • $\begingroup$ Guess you want $x \ge 0$. Otherwise it doesn't hold anyway. $\endgroup$ – Macavity Mar 15 '15 at 6:16
  • $\begingroup$ You are correct. I edited the description. $\endgroup$ – thelionkingrafiki Mar 15 '15 at 12:44
  • $\begingroup$ 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. $\endgroup$ – Stromael Mar 16 '15 at 14:01
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    $\begingroup$ 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/… $\endgroup$ – thelionkingrafiki Mar 16 '15 at 14:11
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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 $k \ge 2$.

Thus, \begin{align} y_{n-1} + y_{n+1} - 2 \, y_n &= \frac{2}{x^2} + \frac{3\cdot 2 \, (n-1)}{x^3} + \frac{4\cdot 3 \,(n-1)(n-2)}{x^4} + \cdots + \frac{(n+1)!}{x^{n+1}}\\ &\le \frac{2}{x^2}\left( 1 + \frac{2^2 \, (n-1)}{x} + \frac{3^2 \,(n-1)(n-2)}{x^2} + \cdots + \frac{n^2 (n-1)!}{x^{n-1}} \right)\\ &= \frac{2}{x^2}\left( 1^2 \, b_1 + 2^2 \, b_2 + 3^2 \, b_2 + \cdots + n^2 \, b_n \right) \\ 1 + 2 \, y_{n-1} - y_n &= 1 + \frac{n-2}{x} + \frac{(n-1)(n-4)}{x^2} + \frac{(n-1)(n-2)(n-6)}{x^3} + \cdots - \frac{n!}{x^{n}}\\ &< 1 + \frac{n-1}{x} + \frac{(n-1)(n-2)}{x^2} + \frac{(n-1)(n-2)(n-3)}{x^3} + \cdots + \frac{(n-1)!}{x^{n-1}}\\ &= b_1 + b_2 + b_3 + \cdots + b_n. \end{align}

Finally, by the Cauchy-Schwarz inequality, we have \begin{align} (1 + 2 \, y_{n-1} - y_n)(y_{n-1} + y_{n+1} - 2 \, y_n) &< \frac{2}{x^2} \left( \sum_{k=1}^{n} k^2 \, b_k \right) \left( \sum_{k=1}^{n} b_k \right) \\ &\le \frac{2}{x^2} \left( \sum_{k=1}^{n} k \, b_k \right)^2 \\ &= 2 \, \left(y_n - y_{n-1} \right)^2. \end{align} Q. E. D.

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