I encountered the following inequality in A Basic Course in Real Analysis by Kumaresan and Ajit Kumar, page 45. I am unable to get a proof for this inequality.
Given,
$x_n = 1 + \sum\limits_{k=1}^{n} \frac{n!}{k! (n-k)!} n^{-k}$
$y_n = \sum\limits_{k=0}^{n} \frac{1}{k!}$
Prove that
If $n,m \in \mathbb{N} $ and $n > m$, then $x_n > y_m$
Edit: The inequality was taken as an obvious result as shown in this picture: Portion from the textbook
The way that $\left(1+\frac1n\right)^n$ is written, seems to indicate that the author wants us to compare the sums term by term. $$ \begin{align} \frac{n!}{k!(n-k)!}n^{-k} &=\frac1{k!}\frac{n!}{n^k(n-k)!}\\ &=\frac1{k!}\prod_{j=0}^{k-1}\frac{n-j}n\\ &\le\frac1{k!} \end{align} $$ The inequality seems to be backward.
Indeed, since $$ \begin{align} \sum_{k=n+1}^\infty\frac1{k!} &=\frac1{(n+1)!}\sum_{k=0}^\infty\frac1{(n+2)^{\overline{k}}}\\ &\le\frac1{(n+1)!}\sum_{k=0}^\infty\frac1{(n+2)^k}\\ &=\frac1{(n+1)!}\frac{n+2}{n+1} \end{align} $$ we have $$ \bbox[5px,border:2px solid #C0A000]{y_n\ge e-\frac1{(n+1)!}\frac{n+2}{n+1}} $$ Furthermore $$ \begin{align} n\log\left(1+\frac1n\right) &=-n\log\left(1-\frac1{n+1}\right)\\ &=n\sum_{k=1}^\infty\frac1{k(n+1)^k}\\ &=\left(1-\frac1{n+1}\right)\sum_{k=1}^\infty\frac1{k(n+1)^{k-1}}\\ &=1-\sum_{k=1}^\infty\frac1{k(k+1)(n+1)^k} \end{align} $$ whereas $$ \begin{align} 1+\log\left(\frac{2n+1}{2n+2}\right) &=1+\log\left(1-\frac1{2n+2}\right)\\ &=1-\sum_{k=1}^\infty\frac1{k2^k(n+1)^k} \end{align} $$ Thus, $$ \bbox[5px,border:2px solid #C0A000]{x_n\le e-\frac e{2n+2}} $$ Therefore, we need $n\gt\frac e2(m+1)!\frac{m+1}{m+2}-1$ to get $x_n\gt y_m$.
For example, $x_n\gt y_5\implies n\ge841$ (formula above gives $838$), and $x_n\gt y_6\implies n\ge6006$ (formula above gives $5993$).
