Study of the sequence $\int_0^n \left(1+\frac{x}{n}\right)^n e^{-\alpha x}\mathrm{d}x$

Question

I wish to understand the following sequence: $$b_n:=\int_0^n \left(1+\frac{x}{n}\right)^n e^{-\alpha x}\mathrm{d}x,$$ where $\alpha$ is a real number and the integral here is Lebesgue's (although I don't know if this will be useful).

I already proved that it is increasing (that is $b_{n+1}\geq b_n$) and that

• If $\alpha>1$, then $b_n\leq \frac{1}{\alpha -1}$. This implies the convergence of $b_n$ but I still don't know the limit.
• If $\alpha<-1$, I proved that $b_n$ diverges.

Obviously, if $\alpha=0$, then $(b_n)$ diverges. However, if $\alpha\in [-1,1]\setminus\{0\}$ I still don't know the behaviour of $(b_n)$ and, even when it converges, I don't know its limit.

Answer 1

One of the main difficulties here is that both the integrand and the interval are changing. We need to reduce the number of things going on.

So instead of thinking about the bounds of the integral changing, think of this as integrating against a characteristic function that changes. That is, write

$$b_n = \int_0^{\infty} \chi_{[0, n]}(x)\left(1 + \frac x n \right)^n e^{-\alpha x} \, dx$$

Now study the integrand as a function of $n$, and invoke monotone convergence. You should find that

$$b_n \to \int_0^{\infty} e^x \cdot e^{-\alpha x} \, dx$$

which is finite if and only if $\alpha > 1$.

Answer 2

Hint: Let $f_n(x)=\chi_{[0,n]}(x)(1+x/n)^ne^{-\alpha x}.$ Then $0\le f_1\le f_2 \le \dots $ on $[0,\infty),$ and the pointwise limit of $f_n(x)$ there is $e^xe^{-\alpha x}.$