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I want to show that:

$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^\infty \frac{1}{k!}.$$

By the binomial theorem

$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \lim_{n\to\infty}\sum_{k=0}^n \frac{1}{k!} \frac{n(n-1)\cdots(n-k+1)}{n^k}=\lim_{n\to\infty}\sum_{k=0}^n f_n(k).$$

Here I have problems with limits, since the limit of the sum depends on $n$ and also the summation terms. I tried to fix it with the following:

Let $g_n:=f_n\chi_{P_n}\colon \mathbb{N}\to \mathbb{R}$ measurable functions in the counting measure $\mu$ in $\mathbb{N}$ we have that $$g_n(k)=f_n(k)\chi_{P_n}(k)\to \frac{1}{k!}$$ for each $k\in\mathbb{N}$ and $P_n=\{1,\ldots,n\}$. But $$|g_n(k)|\leq \frac{1}{k!}:=f(k)\quad\text{and}\quad \int_{\mathbb{N}}f(k)d\mu(k)=\sum_{k=0}^\infty \frac{1}{k!}<3<\infty.$$

So by the Lebesgue Dominated Convergence Theorem,

\begin{align} \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n&=\lim_{n\to\infty}\sum_{k=0}^n f_n(k)\\ &=\lim_{n\to\infty}\int_{P_n}f_n(k)d\mu(k)\\ &=\lim_{n\to\infty}\int_{\mathbb{N}}\chi_{P_n}(k) f_n(k)d\mu(k)\\ &=\lim_{n\to\infty}\int_{\mathbb{N}}g_n(k)d\mu(k)\\ &=\int_{\mathbb{N}}\lim_{n\to\infty}g_n(k)d\mu(k)\\ &=\int_{\mathbb{N}}f(k)d\mu(k)\\ &=\int_{\mathbb{N}}\frac{1}{k!}d\mu(k)\\ &=\sum_{k=0}^\infty \frac{1}{k!}. \end{align}

Am I right?

Davide Giraudo
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Gaston Burrull
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    Aside from the name of the theorem you use and some minor typos, your proof looks fine. I don't think the series version of DCT is called "Lebesgue dominated convergence theorem", some authors just call the series version of DCT "Dominated convergence theorem for series". – achille hui Sep 30 '13 at 05:15

1 Answers1

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Your second and third equalities in the bottom chain are incorrect, they should be $$ = \lim \int_{P_n} f_n d\mu = \lim \int_{\mathbb{N}} \chi_{P_n} f_n d\mu $$ But the rest of the proof goes through if you correct this small typo. Proof looks good.

nullUser
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