Proof of $\left|\sum_{k=1}^N\dfrac{k!}{N^k}-\sum_{k=1}^N\frac{1}{kN}\right|\le\dfrac{1}{4}$ Is it possible to prove the following inequality?
$$\left|\sum_{k=1}^N\dfrac{k!}{N^k}-\sum_{k=1}^N\frac{1}{kN}\right|\le\dfrac{1}{4}$$
Thanks

Yes, it's possible. First use triangle inequality:
$\displaystyle |\sum_{k=1}^N\dfrac{k!}{N^k}-\sum_{k=1}^N\frac{1}{kN}| \le  |\sum_{k=31}^{N}\frac{k!}{N^k}|+|\sum_{k=31}^{N}\frac{1}{kN}|+\sum_{k=1}^{30} \frac{k!}{N^{k}}+\frac{1}{kN}$
1)$\sum_{k=1}^{30} \frac{k!}{N^{k}}+\frac{1}{kN}$ is small when $N$ is large.
2)$\displaystyle |\sum_{k=31}^{N}\frac{k!}{N^k}| \le |\sum_{k=31}^{N}\frac{k!}{k^k}| \le |\sum_{k=31}^{\infty}\frac{k!}{k^k}|\le |\sum_{k=31}^{\infty}\frac{2}{k^2}|$
Last inequality is from $\displaystyle \frac{n!}{n^n} \le \frac{2}{n^2}$ (it's very easy to show).
3)$\displaystyle |\sum_{k=31}^{N}\frac{1}{kN}| \le |\sum_{k=31}^{\infty}\frac{1}{k^2}| $
Series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, so $\sum_{n=30}^{\infty} \frac{1}{n^2}$ is small (see Wolfram).
