From exercise 2.35 of "Measure theory and probability theory" by Krishna and Soumendra:
Using the DCT or otherwise show that for any sequence of real numbers $\{x_n\}$ with $\displaystyle{\lim_{n \to\infty}}x_n=x$:
\begin{gather*}
\displaystyle{\lim_{n \to\infty}}\left(1 + \frac{x_n}n \right)^n=\sum_{j=0}^\infty\frac{x^j}{j!}
\end{gather*}
The book references the precedent exercise where the following is established:
Let $A=((a_{ij}))$ be an infinite matrix of real numbers. Suppose that $\displaystyle{\lim_{i \to\infty}}a_{ij}=a_j \in \mathbb{R}$ and $\displaystyle{\sup_{i}}\left|a_{ij}\right| = b_j \in\mathbb{R}$ for each $j$ and $\sum_{j=0}^\infty b_j < \infty$ then:
\begin{gather*}
\displaystyle{\lim_{n \to\infty}}\sum_{j=1}^\infty \left| a_{ij}-a_j\right|=0
\end{gather*}
That is an application of the DCT with the measurable space $(\mathbb{N}, \mathcal{P}(\mathbb{N}))$ and the counting measure, $a_i(j)$ convergents to $a(j)$ and dominated by the integrable function $b(j)$.
I'm interested in clarifying the computation for casting the first problem as an istance of the second, thus solving it by the DCT. Any help would be appreciated, thanks.
Best Answer
The DCT basically lets you swap limits and integrals (or in this case, sums). Since we have a limit on the left, and a sum on the right, a strategy presents itself -- Find a way to write the stuff inside the limit as a sum, then swap the order (by DCT), then evaluate each limit. Let's see a sketch of this plan in action:
Notice (by the binomial theorem) we can write the left side as a limit of a sum
$$ \lim_{n \to \infty} \left ( 1 + \frac{x_n}{n} \right )^n = \lim_{n \to \infty} \sum_{i = 0}^n \binom{n}{i} \frac{x_n^i}{n^i} $$
We want to apply DCT, so we need to fix the "bounds of integration" so they no longer depend on $n$. We can do that by using a characteristic function, then "integrating" over all the naturals
$$ \lim_{n \to \infty} \sum_{i = 0}^n \binom{n}{i} \frac{x_n^i}{n^i} = \lim_{n \to \infty} \sum_{i = 0}^\infty \binom{n}{i} \frac{x_n^i}{n^i} \chi_{i \leq n} $$
Then (by DCT) we can swap the order of the sum and the limit (you should rigorously show the hypotheses hold)
$$ \lim_{n \to \infty} \sum_{i = 0}^\infty \binom{n}{i} \frac{x_n^i}{n^i} \chi_{i \leq n} = \sum_{i = 0}^\infty \lim_{n \to \infty} \binom{n}{i} \frac{x_n^i}{n^i} \chi_{i \leq n} $$
Now we can focus attention on any given term of the right hand side
$$ \lim_{n \to \infty} \binom{n}{i} \frac{x_n^i}{n^i} \chi_{i \leq n} = \lim_{n \to \infty} \frac{n! x_n^i \chi_{i \leq n}}{(n-i)! i! n^i} $$
We do some simple rewriting
$$ \lim_{n \to \infty} \frac{n! x_n^i \chi_{i \leq n}}{(n-i)! i! n^i} = \lim_{n \to \infty} \frac{\frac{n!}{(n-i)!}}{n^i} \frac{x_n^i}{i!} \chi_{i \leq n} $$
Now taking limits (again, I leave it to you to work out the details),
$$\frac{\frac{n!}{(n-i)!}}{n^i} \to 1$$ $$\chi_{i \leq n} \to 1$$ $$\frac{x_n^i}{i!} \to \frac{x^i}{i!}$$
So we see (for each term)
$$ \lim_{n \to \infty} \frac{\frac{n!}{(n-i)!}}{n^i} \frac{x_n^i}{i!} \chi_{i \leq n} = \frac{x^i}{i!} $$
Where did we start, and where did we end?
$$ \lim_{n \to \infty} \left ( 1 + \frac{x_n}{n} \right )^n = \sum_{i=0}^\infty \frac{x^i}{i!} $$
as desired.
I hope this helps ^_^