$\sum_{k=0}^{n} \frac{1}{k!}\leq \left(1+\frac{1}{n}\right)^{n+\alpha} $ 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}
$\forall n\in\mathbb{N}$.
 A: 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.
