$[(1+\delta)e^{-\delta}]^{t-1} \int_{\delta}^{\infty}(z+1)e^{-z}dz \leqslant 2\exp{ \{1-\frac{t \delta^2}{2} + \frac{\delta^3}{3(1-\delta)} \}}$

Question

I am working on the exercise 1.3.19 proving Stirling's formula in Stroock's "Probability Theory: an Analytic View". I need help in proving the following inequality:

\begin{equation*} [(1+\delta)e^{-\delta}]^{t-1} \int_{\delta}^{\infty}(z+1)e^{-z}dz \leqslant 2\exp{ \{1-\frac{t \delta^2}{2} + \frac{\delta^3}{3(1-\delta)} \}} \end{equation*}

Here $t \in \mathbb{R}$ is assumed to be large while $\delta \in (0,1)$. The inequality arises from estimating a part of gamma function after a change of variable.

First of all, the integral can be solved:

\begin{equation*} \int_{\delta}^{\infty}(z+1)e^{-z}dz = (\delta+2)e^{-\delta} \end{equation*}

Combine the terms, the left hand side is:

\begin{equation*} [(1+\delta)]^{t-1} (\delta+2) e^{-t\delta} \end{equation*}

It seems the right hand side might from some sort of Taylor expansion. However, the derivative seems complicated and is not giving me what I want. I am stuck here.

Answer 1

You have to exponentials on both side, so your first instinct was not to simplify them ?

$$(1+\delta)^{t-1}(2+\delta)e^{-t\delta}\leq 2e^{1-t\delta^2/2 + \delta^3/(3(1-\delta))}\iff$$ $$(1+\delta)^{t-1}(1+\delta/2)\leq e^{1+t\delta-t\delta^2/2 + \delta^3/(3(1-\delta))}\iff$$ $$(t-1)\ln(1+\delta) + \ln\left(1+\frac{\delta}{2}\right)\leq 1+t\left(\delta-\dfrac{\delta^2}{2}\right) + \dfrac{\delta^3}{3(1-\delta)}\iff $$ $$\left(\ln(1+\delta)-\delta+\dfrac{\delta^2}{2}\right)t\leq 1+ \ln\dfrac{1+\delta}{1+\frac{\delta}{2}} + \dfrac{\delta^3}{3(1-\delta)}\iff$$ $$t\leq\dfrac{1+ \ln\dfrac{1+\delta}{1+\frac{\delta}{2}} + \dfrac{\delta^3}{3(1-\delta)}}{\left(\ln(1+\delta)-\delta+\dfrac{\delta^2}{2}\right)}.$$

So unfortunately, your inequality as stated is not globally true although the above upper bound does appear to be "large" since the denominator will be very small in $(0,1).$ For example, when $\delta = 0.5$, you get $t \leq 41.5...$