$\sum_{k=0}^{n} \frac{1}{k!}\leq \left(1+\frac{1}{n}\right)^{n+\alpha} $

Question

Find a value $\alpha\in\mathbb{N}$ such that \begin{equation} \sum_{k=0}^{n} \frac{1}{k!}\leq \left(1+\frac{1}{n}\right)^{n+\alpha} \end{equation} $\forall n\in\mathbb{N}$.

Answer 1

Let's solve $\left(1+\frac1n\right)^{n+\alpha}\ge e$

$$ (n+\alpha)\log\left(1+\frac1n\right)\ge1 $$ Note that $\log(1-x)=-\int \frac1{1-x}\text dx=-\int\sum_{k=0}^\infty x^k\text dx=-\sum_{k=1}^\infty \frac{x^k}{k}$

So we want $$\begin{align}-(n+\alpha)\sum_{k=1}^\infty \frac{(-1)^k}{kn^k}&\ge 1\\ n\sum_{k=1}^\infty \frac{(-1)^{k+1}}{kn^{k}}+\alpha\sum_{k=1}^\infty \frac{(-1)^{k+1}}{kn^{k}}&\ge 1\\ \sum_{k=1}^\infty \frac{(-1)^k}{(k+1)n^k}-\alpha\sum_{k=1}^\infty \frac{(-1)^k}{kn^k}&\ge 0\\ \end{align}$$

Note that this is an alternating series, so each term provides an upper or lower bound. The first few terms are $$\left(\alpha-\frac12\right)n^{-1}+\left(\frac13-\frac12\alpha\right)n^{-2}+\left(\frac13\alpha-\frac14\right)n^{-3}+\ldots$$

So if $\alpha=\frac12$, we see the sequence is bounded below by $0$ and bounded above by $\frac1{12n^2}$. Thus $\alpha =\frac12$ not only gives a sequence that is greater than $e$ for all $n$, but it also gives the fastest convergence to $e$ ($O(n^{-2})$ as opposed to $O(n^{-1})$). Obviously $\alpha\ge\frac12$ satisfy your inequality.