Equivalent of a sequence defined as induction formula

Question

Let $(a_n)_{n \in \mathbb{N}}$ defined as $$ a_{n+1}=a_n+\frac{a_{n-1}}{n+1} \text{ with } a_0=a_1=1 $$ I've shown that for $n \in \mathbb{N}$ $$ a_n=\sum_{k=0}^{n}\frac{n-k+1}{k!}\left(-1\right)^k $$ How can I show, rigorously, that $$ a_n \underset{(+\infty)}{\sim}\frac{n}{e} \ ? $$ I know that if a series $\sum_{n \geq 0}b_n$ diverges and that $b_n \underset{(+\infty)}{\sim}c_n$ then $$ \sum_{k=0}^{n}b_k \underset{(+\infty)}{\sim}\sum_{k=0}^{n}c_k $$ can I use this result to prove it ?

Answer 1

Simply note that $$\sum_{k=0}^{n}\frac{n-k+1}{k!}\left(-1\right)^k = n\sum_{k=0}^n \frac{(-1)^k}{k!} - \sum_{k=0}^n\frac{k-1}{k!}(-1)^k$$

$\displaystyle \sum_{k=0}^n \frac{(-1)^k}{k!}$ converges to $e^{-1}$ and $\displaystyle \sum_{k=0}^n\frac{k-1}{k!}(-1)^k$ converges absolutely to some irrelevant value.

Thus $\displaystyle \sum_{k=0}^{n}\frac{n-k+1}{k!}\left(-1\right)^k =n(e^{-1}+o(1))+O(1)=ne^{-1} +o(n)$.