Suppose $b \gt a \gt 0$ Prove that $\lim_{n \to \infty} \int_{a}^{b} (1+ \frac{x}{n})^n e^{-x}dx=b-a$

Question

I am attempting to prove this. My thought process is to let $f_n(x)= (1+ \frac{x}{n})^n e^{-x}$ and show this converges uniformly to the function $f(x)=1$. then from theorems in my text book i have that $\lim_{n \to \infty} \int_{a}^{b} (1+ \frac{x}{n})^n e^{-x}dx = \int_{a}^{b} \lim_{n \to \infty}((1+ \frac{x}{n})^n e^{-x})dx$ which would be $b-a$. However I am struggling to show uniform convergence of $f_n$.

Answer 1

Hint. Take the exp-log form :

$$(1+x/n)^n e^{-x} = \exp \{n\ln(1+x/n)-x\} $$

so all you have to do is just prove uniform convergence of $n\ln(1+x/n)$ towards $x$, uniformly for every $x$ in $[a,b]$.

Answer 2

I suggest that you use Dini's theorem to conclude uniform convergence. Your sequence $f_n$ is monotone and continuous, and you have pointwise convergence to the continuous function $1$.

Answer 3

As told by Tlon Uqbar Orbis Tertius,

$$(1+x/n)^n e^{-x} = \exp \{n\ln(1+x/n)-x\} $$

As $n \to +\infty$ and noting that $x$ is bounded $(0 < x \leq b)$, there exists some $N$ so that for all $n \geq N$ we know that $x/n < 1$.

It is well-known that if $0 < y < 1$ than

$$\displaystyle\ln(1+y) = y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\dots < y$$

(cfr. convergence of alternating series by Leibniz test)

$$\displaystyle\ln\left(1+\frac{x}{n}\right) < \frac{x}{n}$$ $$\displaystyle n.\ln\left(1+\frac{x}{n}\right) < x $$ $$\left(1+\frac{x}{n} \right)^n < e^x $$