Establishing an identity for exponential function through an application of DCT 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.
 A: 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 ^_^
