A basic exercise about Lebesgue's Dominated Convergence Theorem

Question

Consider the sequence $g_n(x)=(1-\frac{x}{n})^n$ for $x\in[-1,1]$. Calculate the value $$\lim_{n\to\infty}\int_{-1}^1g_n(x)dx$$ by showing the existence of the limit with the aid of the Lebesgue's dominated convergence theorem (LDCT).

Attempt. For $n\in\Bbb N,x\in[-1,1]$, note that $|g_n(x)|=\left(1-\frac{x}{n}\right)^n\leq 2$ and the function $h(x)=2$ is integrable on $[-1,1]$. So, we may apply (LDCT) as follows: $$\lim_{n\to \infty}\int_{-1}^1g_n(x)dx=\int_{-1}^1\lim_{n\to \infty}g_n(x)dx=\int_{-1}^1 1dx=2.$$ I would be glad if someone could check my attempt. Thanks!

Answer 1

Hint. Note that the point-wise limit of sequence of functions $(g_n(x))_n$ is $g(x)=e^{-x}$ (it is not $1$). Hence, by following your approach, we may apply the Lebesgue's dominated convergence theorem (note that $|g_n(x)|$ is bounded by $e$) and $$\lim_{n\to \infty}\int_{-1}^1g_n(x)dx=\int_{-1}^1 e^{-x}dx.$$ What is the final result?

Answer 2

It's already been pointed out that the limit of $g_n$ is not what you said. Your inequality $|g_n|\le 2$ is also wrong.

To get a correct bound suitable for applying DCT, "recall" that $$\log(1+t)\le t\quad(t>-1).$$

Answer 3

I would like to write the complete solution to here by using the answer of Robert Z. We know that the point-wise limit of the sequence $(g_n)$ is $g(x)=e^{-x}$ and we show that $|g_n(x)|<e$ for all $n\in\Bbb N, x\in[-1,1]$ to apply LDCT. First, note that $$\binom{n}{k}\frac{1}{n^k}=\frac{n!}{(n-k)!k!}\frac{1}{n^k}=\frac{n(n-1)\dots(n-k+1)}{k!}\frac{1}{n^k}\leq \frac{1}{k!}.$$ So, for any $n\in\Bbb N, x\in[-1,1]$, we have $$|g_n(x)|=\left|\left(1-\frac{x}{n}\right)^n\right|=\left(1-\frac{x}{n}\right)^n\leq \left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}1^{n-k}\left(\frac{1}{n}\right)^k\leq \sum_{k=0}^n\frac{1}{k!}<\sum_{k=0}^\infty\frac{1}{k!}=e.$$ By LDCT, we get $$\lim_{n\to \infty}\int_{-1}^1g_n(x)dx=\int_{-1}^1\lim_{n\to \infty}g_n(x)dx=\int_{-1}^1 e^{-x}dx=e-\frac{1}{e}.$$