When I was studying the dominated convergence theorem, I was glad to come across this problem:
Use the dominated convergence theorem (DCT) to show that $$\lim \int_{0}^{n} \left(1-\frac{x}{n} \right)^{n}x^sdx=\int_{0}^{\infty}e^{-x}x^sdx$$ where $s<-1$.
I let $\displaystyle f_n(x)= \left(1-\frac{x}{n} \right)^{n}$, clearly $\lim f_n(x)=e^{-x}$. And $e^{-x}$ is integrable.
The book I am using stated the dominated convergence theorem as follows:
Theorem: Let $f_n$ be a sequence of measurable functions such that $f_n \to f$ almost everywhere. If there exist a real valued function $g$ define on a measure space such that for each $n$, $|f_n| \le g$ almost everywhere then $f$ is integrable and $$\int f=\lim\int f_n$$
To be able to use the DCT it necessary to find any real valued function $g$ that dominate $f_n$. That is where I get stuck, can somebody help me out?
EDIT:
The $\lim$ here means limit as n tend to infinity.
Best Answer
Let me write out my comment so that this question gets an answer. The dominated convergence theorem says the following.
If a sequence of measurable functions $\displaystyle \{f_n(x)\}_{n=1}^{\infty}$, on the measure space $(\Omega, \mathcal{F}, \mu)$, converge point-wise to a function $f(x)$ and every $f_n(x)$ is dominated by some integrable function $g(x)$ i.e. $\lvert f_n(x) \rvert \leq g(x)$, $\forall n$, $\forall x \in \Omega$ and $\displaystyle \int_{\Omega} g d\mu < \infty $, then $\displaystyle \int_{\Omega} f d \mu$ exists and $$\lim_{n \rightarrow \infty} \int_{\Omega} f_n d \mu = \int_{\Omega} f d \mu$$
Note that it is important that the dominating function $g$ is integrable as Henry T. Horton points out in the comments. Also, the condition $\lvert f_n \rvert \leq g$ everywhere, can be relaxed to $\lvert f_n \rvert \leq g$ $\mu$-almost everywhere.
Also, we need $s > -1$. Otherwise the integrals given in the question diverge.
Now lets apply this to the problem at hand. We want to evaluate $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx$.
Define $$f_n(x) = \begin{cases} \left( 1 - \frac{x}n\right)^n x^s & \text{ if }x \in [0, n]\\ 0 & \text{ otherwise}\end{cases}$$
Hence, $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx$.
Now note that $f_n(x)$ converges point-wise to $e^{-x} x^s$ on $[0, \infty)$. This is so since $$\displaystyle \lim_{n \rightarrow \infty} \left(1 - \frac{x}{n} \right)^n = \exp(-x).$$
More importantly, $f_n(x)$ is dominated by $e^{-x} x^s$ on $[0,\infty)$ i.e. $\lvert f_n(x) \rvert \leq e^{-x} x^s$. This follows from the fact that $$1 - t \leq \exp(-t)$$ whenever $0 \leq t \leq 1$. Hence, we get that $$\left(1 - \frac{x}{n} \right) \leq \exp \left(-\frac{x}{n} \right)$$ which in-turn gives us $$\left(1 - \frac{x}{n} \right)^n \leq \exp \left(-x \right).$$ Hence, $\displaystyle f_n(x) \leq \exp(-x) x^s $. Also, $\displaystyle \int_0^{\infty} \exp(-x) x^s dx < \infty$ for all $s > -1$. Hence, in this case, the dominating function is the same as the limit function.
Putting all these things together, we get the desired result.
\begin{align} \displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx & = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx & \text{(From the definition of $f_n(x)$)}\\ & = \int_0^{\infty} \lim_{n \rightarrow \infty} f_n(x) dx & \text{(Since $f_n(x)$ is dominated by $f(x)$)}\\ & = \int_0^{\infty} f(x) dx & \text{(Since $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = f(x)$)}\\ & = \int_0^{\infty} \exp(-x) x^s dx \end{align}