I know that $$\lim_{n\to\infty}\left(1+\dfrac{1}{n^2}\right)^n = 1$$ but by applyling Tannery's theorem, I am getting the answer as zero. Can somebody, help me find the error in the solution below?
$$S_n = \Bigg(1+\dfrac{1}{n^2}\Bigg)^n = \sum_{k=0}^n{n \choose k}\bigg(\frac{1}{n^2}\bigg)^k = \sum_{k=0}^n\bigg(\frac{n}{n}\bigg)\bigg(\frac{n-1}{n}\bigg)\cdots\bigg(\frac{n-k+1}{n}\bigg)\bigg(\frac{1}{n^kk!}\bigg)$$
$$\implies S_n = \sum_{k=0}^\infty a_k(n) \text{ , where } a_k(n) = \begin{cases} 0 & \text{ if } k>n \\ \bigg(\frac{n}{n}\bigg)\bigg(\frac{n-1}{n}\bigg)\cdots\bigg(\frac{n-k+1}{n}\bigg)\bigg(\frac{1}{n^kk!}\bigg) & \text{ if }k\le n \end{cases} $$
clearly, $a_k(n)\stackrel{n\to\infty}{\longrightarrow}0 = b_k \text{ ,} \forall k$
and $|a_k(n)| \le \bigg(\frac{n}{n}\bigg)\bigg(\frac{n-1}{n}\bigg)\cdots\bigg(\frac{n-k+1}{n}\bigg)\bigg(\frac{1}{k!}\bigg)\le \frac{1}{k!},\forall n$
and $\sum \frac{1}{k!} = e< \infty$
$$\implies \text{ by Tannery's theorem } \lim_{n\to\infty} S_n = \lim_{n\to\infty}\sum_{k=0}^\infty a_k(n) = \sum_{k=0}^\infty \lim_{n\to\infty} a_k(n) = \sum_{k=0}^\infty b_k = 0$$
But the correct answer is 1. So, where did I go wrong?
The error is in the definition of $a_k(n)$, since for $k=0$ we have $$ a_k(n) = {n \choose 0} \left(\frac{1}{n^2}\right)^0 = 1. $$ This makes the following statement wrong: $$ a_k(n) {\to} 0 = b_k, \forall k $$ since is not true for $k=0$, as for $k=0$ we have $a_0(n) = 1$, $\forall n$. When you substitute it with $$ b_k = \begin{cases} 1 & \text{if } k=0 \\ 0 & \text{if } k \geq 1 \end{cases}, $$ you will obtain $\lim_{n\to\infty}S_n = \sum_{k=0}^\infty b_k = 1$.