Subsequences of convergent series

Question

We know that all subsequences $x_{\phi(n)}$ of a convergent sequences $x_n$ have the same limit : $$\lim_{n \to \infty} x_n = \lim_{n \to \infty} x_{\phi(n)}$$ Now we consider $x_n=\frac{n^3}{n!}$, we have that $S_n = \sum_{k=0}^n x_k = \sum_{k=0}^n\frac{k^3}{k!}$.

Suppose we already know that $\sum_{n=0}^\infty \frac{n^3}{n!}= \lim_{n \to \infty} S_n=5e$.

My question is : as the series converges (absolutely), i.e. $S_n$ has a limit when $n \to \infty$, why does the subsequence $S_{2n} = \sum_{k=0}^n\frac{(2k)^3}{(2k)!}$ seem to have a different limit that is not $5e$ ? I thought we should have $\lim_{n \to \infty} S_n = \lim_{n \to \infty} S_{2n}$ but I also see that $S_{2n}$ is missing half of the terms from the original sum so its limit "must" be lower. So how does subsequences work with series ? I am sure there is something very stupid that I don't get.

Answer 1

Because $$S_{2n} = \sum_{k=1}^n \frac{(2k)^3}{(2k)!}$$ is false. The correct $S_{2n}$ is $$S_{2n} = \sum_{k=1}^{2n} \frac{k^3}{k!}$$

Answer 2

You made a mistake: $$S_{2n} = \sum_{k=0}^{2n} \frac{k^3}{k!}$$ and not:

$$S_{2n} = \sum_{k=0}^{n} \frac{(2k)^3}{(2k)!}$$

Answer 3

You are wrong when you assert that$$S_{2n}=\sum_{k=0}^n\frac{(2k)^3}{(2k)!}.$$In fact,$$S_{2n}=\sum_{k=0}^{2n}\frac{k^3}{k!},$$which is a different thing. And there is no reason for you to assume that$$\lim_{n\to\infty}\sum_{k=0}^n\frac{(2k)^3}{(2k)!}=\lim_{n\to\infty}\sum_{k=0}^{2n}\frac{k^3}{k!}.$$