Find the limit of $\lim_{x\to \infty}(\frac{x}{x})^x+(\frac{x-1}{x})^x+(\frac{x-2}{x})^x......+(\frac{1}{x})^x$ Find the limit of $\lim_{x\to \infty}(\frac{x}{x})^x+(\frac{x-1}{x})^x+(\frac{x-2}{x})^x......+(\frac{1}{x})^x$

$\lim_{x\to \infty}(\frac{x}{x})^x+(\frac{x-1}{x})^x+(\frac{x-2}{x})^x......+(\frac{1}{x})^x$
$=\lim_{x\to \infty}\frac{1^x+2^x+3^x+.....+x^x}{x^x}$
This is in $\frac{\infty}{\infty}$ form ,so i applied L hospital rule.
$=\lim_{x\to \infty}\frac{2^x\log 2+3^x\log 3+.....+x^x(1+\log x)}{x^x(1+\log x)}$
But i am stuck here and could not solve further.Please help.

Note that
$$
\begin{align}
\lim_{x\to\infty}\left(\frac{x-k}x\right)^x
&=\lim_{x\to\infty}\left(1-\frac kx\right)^x\\
&=e^{-k}
\end{align}
$$
Therefore, the sum of the limits is
$$
1+e^{-1}+e^{-2}+\dots=\frac1{1-\frac1e}=\frac e{e-1}
$$

First, since the series was given as $\left(\frac xx\right)^x+\cdots+\left(\frac1x\right)^x$, I assumed that $x\in\mathbb{Z}$.
However, even if we don't assume that $x\in\mathbb{Z}$, but that we only include $\left(\frac{x-k}x\right)^x$ for $k\lt x$, Bernoulli's Inequality says that $\left(1-\frac kx\right)^x$ is increasing in $x$. That is, for $y\ge x\ge k$,
$$
\overbrace{\left(1-\frac ky\right)^{\large\frac yx}\ge1-\frac yx\frac ky}^{\text{Bernoulli's Inequality}}
\implies
\overbrace{\left(1-\frac ky\right)^y\ge\left(1-\frac kx\right)^{x\vphantom{\large\frac yx}}}^{\text{raising to the $x$ power}}
$$
Thus, we only need invoke Monotone Convergence (the terms for $k\ge x$ are $0$) to assure that the limit of the sum is the sum of the limits.
