For what values of $p>0$ is $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^ne^x}{n^p}dx=0$?

Question

For what values of $p>0$ is $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^ne^x}{n^p}dx=0$?

My thoughts: We know that $(1-\frac{x}{n})^n\leq e^x$, so the numerator is $\leq e^{2x}$. So, we can play with $\frac{e^{2x}}{n^p}$. From here, I am not quite sure what to do. I would really like to be able to find the supremum over $n$, but I can't really minimize the denominator to be able to replace $n$ with something in terms of $x$, because I only have $n^p$ down there. On the other hand, I feel like I should be splitting up the integral from $0$ to $1$ and then from $1$ to $\infty$ based on the denominator. Then, fix some $x$, and just use $p$ integral properties to get that $p\in(0,1)$, but I am not quite sure. Any help is greatly appreciated!

A quick edit: I realize that I made a big mistake above by overlooking the minus sign, so instead the integrand is bounded above by $\frac{1}{n^p}\leq 1$ as $n\rightarrow \infty$, and so we can use DCT and then treat it like a $p$ integral.

A second edit: For Sangchul Lee, I edited the integral to make the upper bound $n$ so he can expand on how he got his approximation. Thank you!

Answer 1

To answer the question, it is sufficient to show:

$$ \color{navy}{ \lim_{n\to\infty} \frac{1}{\sqrt{n}} \int_{0}^{n} \left(1-\frac{x}{n}\right)^n e^x \, \mathrm{d}x = \sqrt{\frac{\pi}{2}} \qquad\text{as}\quad n\to\infty. } $$

Indeed, substitute $x = \sqrt{n}u$ to find that

\begin{align*} \frac{1}{\sqrt{n}} \int_{0}^{n} \left(1-\frac{x}{n}\right)^n e^x \, \mathrm{d}x &= \int_{0}^{\sqrt{n}} \Bigl(1 - \frac{u}{\sqrt{n}}\Bigr)^n e^{\sqrt{n}u} \, \mathrm{d}u \\ &= \int_{0}^{\infty} \Bigl(1 - \frac{u}{\sqrt{n}}\Bigr)^n e^{\sqrt{n}u} \mathbf{1}_{[0, \sqrt{n})}(u) \, \mathrm{d}u \end{align*}

To analyze the integral in the right-hand side, we note that

$$ 1 - t = e^{\log(1-t)} = e^{-\sum_{k=1}^{\infty} t^k/k} \leq e^{-t-t^2/2} $$

holds for $ t \in [0, 1)$. Using this, we find that

$$ \Bigl(1 - \frac{u}{\sqrt{n}}\Bigr)^n e^{\sqrt{n}u} \mathbf{1}_{[0, \sqrt{n})}(u) \leq e^{-u^2/2} $$

for $u \geq 0$. It is a easy to check that $$\lim_{n\rightarrow\infty}\Bigl(1 - \frac{u}{\sqrt{n}}\Bigr)^n e^{\sqrt{n}u} \mathbf{1}_{[0, \sqrt{n})}(u)=e^{-u^2/2}$$

So by the dominated convergence theorem,

\begin{align*} \lim_{n\to\infty} \frac{1}{\sqrt{n}} \int_{0}^{n} \left(1-\frac{x}{n}\right)^n e^x \, \mathrm{d}x &= \int_{0}^{\infty} \lim_{n\to\infty} \Bigl(1 - \frac{u}{\sqrt{n}}\Bigr)^n e^{\sqrt{n}u} \mathbf{1}_{[0, \sqrt{n})}(u) \, \mathrm{d}u \\ &= \int_{0}^{\infty} e^{-u^2/2} \, \mathrm{d}u \\ &= \sqrt{\frac{\pi}{2}}. \end{align*}

Answer 2

If the upper limit is $n$ with $n>0$ $$I_n=\int_{0}^{n}\left(1-\frac{x}{n}\right)^n e^x\,dx=n\int_{0}^{1} (1-t)^n\,e^{n t}\,dt=e^n n^{-n} \Big[\Gamma (n+1)-\Gamma (n+1,n)\Big]$$ $$I_n=e^n n^{-n}\Gamma (n+1)\Big[1-\frac {\Gamma (n+1,n)}{\Gamma (n+1)}\Big]$$

If, for large $n$, we make the approximation $$\Gamma (n+1,n)\sim \Gamma (n+1,n+1)$$ (have a look at $8.11.12$ here) $$\Gamma (n+1,n)\sim \frac{1}{24} e^{-(n+1)} (n+1)^n \left(\frac{\sqrt{2 \pi } (12 n+13)}{\sqrt{n+1}}-8\right)$$ $$\log\Big[\frac {\Gamma (n+1,n)}{\Gamma (n+1)}\Big]=-\log (2)-\frac{1}{3} \sqrt{\frac{2}{\pi n}} -\frac{1}{9 \pi n}+O\left(\frac{1}{n^{3/2}}\right)$$ $$I_n=e^n n^{-n}\Gamma (n+1)\Big[\frac{1}{2}+\frac{1}{3 \sqrt{2 \pi n }}+O\left(\frac{1}{n^{3/2}}\right) \Big]$$ Finishing with Stirling approximation $$I_n=\sqrt{\frac{\pi n}{2}} +\frac{1}{3}+\frac{1}{12} \sqrt{\frac{\pi }{2n}}-\frac{1}{6 n}+O\left(\frac{1}{n^{3/2}}\right) $$

Edit

Using, as @Gary commented $$\Gamma (n + 1,n) = n\Gamma (n,n) + n^n e^{ - n}$$ and the expansion given in 1 $$I_n=\sqrt{\frac{\pi n }{2}} -\frac 23+\frac{1}{12} \sqrt{\frac{\pi }{2n}}+\frac{4}{135 n}+\frac{1}{288} \sqrt{\frac{\pi }{2n^3}} +O\left(\frac{1}{n^{2}}\right) $$ which is almost exact $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 5 & 2.18885 & 2.18867 \\ 10 & 3.33279 & 3.33275 \\ 15 & 4.21642 & 4.21640 \\ 20 & 4.96321 & 4.96320 \\ 25 & 5.62201 & 5.62201 \end{array} \right)$$

Answer 3

Assume that $n \ge 2$. Let $$I_n := \int_0^n \left(1 - \frac{x}{n}\right)^n \mathrm{e}^x \,\mathrm{d} x.$$

Using $\ln(1 + u) \le u$ for all $u > -1$, we have, for all $0 \le x < n$, \begin{align*} \ln\left(1 - \frac{x}{n}\right) &= \ln \left(1 - \frac{\sqrt n}{n}\right) + \ln\left(1 - \left(1 - \frac{\sqrt n}{n}\right)^{-1}\frac{x - \sqrt n}{n}\right)\\[6pt] &\le - \frac{\sqrt n}{n} - \left(1 - \frac{\sqrt n}{n}\right)^{-1}\frac{x - \sqrt n}{n}\\ &= - \frac{x}{n - \sqrt n} + \frac{1}{n - \sqrt n}. \end{align*} Thus, we have \begin{align*} I_n &\le \mathrm{e}^{n/(n - \sqrt n)}\int_0^n \mathrm{e}^{- nx/(n - \sqrt n) + x}\, \mathrm{d} x\\ &= \mathrm{e}^{n/(n - \sqrt n)} (\sqrt n - 1)(1 - \mathrm{e}^{-n/(\sqrt n - 1)})\\ &\le \mathrm{e}^{4}\sqrt n. \end{align*}

On the other hand, we have, for all $x \in [0, \sqrt n)$, $$\left(1 - \frac{x}{n}\right) \mathrm{e}^{x/n} \ge \left(1 - \frac{x}{n}\right)\left(1 + \frac{x}{n}\right) = 1 - \frac{x^2}{n}\cdot \frac{1}{n} \ge \left(1 - \frac{x^2}{n}\right)^{1/n}$$ where we have used $\mathrm{e}^u \ge 1 + u$ for all $u\ge 0$, and Bernoulli inequality $(1 + v)^r \le 1 + vr$ for all $0 < r \le 1$ and $v > -1$. Thus, we have \begin{align*} I_n &\ge \int_0^{\sqrt n} \left(1 - \frac{x}{n}\right)^n \mathrm{e}^x \,\mathrm{d} x\\ &\ge \int_0^{\sqrt n} \left(1 - \frac{x^2}{n}\right) \,\mathrm{d} x\\ &= \frac23 \sqrt n. \end{align*}

Thus, we have $$\frac23 \sqrt n \le I_n \le \mathrm{e}^4 \sqrt{n}, \quad \forall n\ge 2.$$

Thus, $\lim_{n\to \infty} \frac{I_n}{n^p} = 0$ if and only if $p > 1/2$.