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.
 A: For a fixed $m$:
$$\lim_{x\to\infty}\left(\frac{x-m}{x}\right)^x= e^{-m}\tag{1}$$
So if the series converges at all, it must converge to a value at least as big as:
$$\sum_{m=0}^\infty e^{-m} = \frac{1}{1-e^{-1}}=\frac{e}{e-1}$$
Letting $f_m(x)=\left(\frac{x-m}{x}\right)^x$, then define $g_m(x)=\ln f_m(x)=x\left(\log(x-m)-\log(x)\right)$, and you get:
$$g_n'(x)=\log(x-m)-\log(x) + \frac{x}{x-m} - 1=\log\left(1-\frac{m}{x}\right) +\dfrac{\frac{m}{x}}{1-\frac mx}$$
You can use the power series for $\frac{1}{1-z}$ and $\log(1-z)$ to show that this value is positive, and hence $g_m$ is increasing for $x>m$, and this that the above series converges to the hoped-for value.
This shows that $F(x)=\sum_{0}^x f_m(x)$ is strictly increasing, and thus we have:
$$\lim_x F(x)=\sup_x F(x)\leq \sum_{m} e^{-m}$$
And also have shown above that $$\sup_x F(x)\geq \sum_{m} e^{-m}$$
So the limit is equal to $\frac{e}{e-1}$.

Aside: If you have a sequence of sequences, $a_{n,k}$, and for each $k$, you have $\lim_{n\to\infty} a_{n,k}=A_k$, it is not always true that:
$$\lim_{n\to \infty}\sum_{k=0}^\infty a_{n,k} = \sum_{k=0}^\infty A_k$$
A simple example where it isn't true is: $$a_{n,k}=\delta_{n,k}=\begin{cases}1&n=k\\0&n\neq k\end{cases}$$
Then for each $k$, $\lim_{n\to\infty} a_{n,k}=0$, and for all $n$, $\sum_{k=0}^{\infty}a_{n,k}=1$, so the limit of the sum is not the sum of the the limits.
If, for each $k$, however, $a_{1,k},a_{2,k},\dots$ is increasing, then the limit of the sums is the sum of the limit.
A: 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.
A: Edit: robjohn has corrected his answer, but I'll leave my proof of the general theorem here, since the techniques involved are instructive.
$\def\nn{\mathbb{N}}$
Solution
In general the following MCT (monotone convergence theorem) holds:

$\sum_{k=0}^\infty f_n(k) \to \sum_{k=0}^\infty f(k)$ as $n \to \infty$ for any increasing sequence of non-negative functions $(f_n)_{n\in\nn}$ such that $f_n \to f$ pointwise as $n \to \infty$.

In this particular instance $f_n(k) = (1-\frac{k}{n})^n$ for any $n \in \nn$ and $k \in \nn_{\le n}$, and $f_n(k) = 0$ for any $n \in \nn$ and $k \in \nn_{>n}$. To apply MCT you must first prove that $(1-\frac{k}{n})^n \le (1-\frac{k}{n+1})^{n+1}$ for any $n,k \in \nn$, which is easy to see from $(1-\frac{k}{n})^n = \exp(n\ln(1-\frac{k}{n})) = \exp( - \sum_{i=1}^\infty \frac{k^i}{n^{i-1}} )$. Then MCT gives immediately that $\sum_{k=0}^{n-1} (1-\frac{k}{n})^n \to \sum_{k=0}^\infty e^{-k} = \frac{1}{1-e^{-1}}$.
Proof of MCT
Clearly $\sum_{k=0}^\infty f_n(k) \le \sum_{k=0}^\infty f(k)$ for any $n \in \nn$, because $f_n(k) \le f_{n+1}(k)$ for any $n,k \in \nn$.
On the other hand, for any $c \in \nn$, we have $\sum_{k=0}^\infty f_n(k) \ge \sum_{k=0}^c f_n(k) \to \sum_{k=0}^c f(k)$ as $n \to \infty$, where the inequality is by non-negativity of $(f_n)_{n\in\nn}$, and the limit is by their pointwise convergence to $f$ and because we only have a limit of a finite sum. But $\sum_{k=0}^c f(k) \to \sum_{k=0}^\infty f(k)$ by definition of infinite series, where the infinite sum exists (could be $\infty$) by non-negativity. Thus given any $ε > 0$, we can choose $c \in \nn$ such that $\sum_{k=0}^c f(k) > \sum_{k=0}^\infty f(k) - ε$, and then choose $m \in \nn$ such that $\sum_{k=0}^c f_n(k) > \sum_{k=0}^c f(k) - ε$ for any $n \in \nn_{\ge m}$. This implies that $\sum_{k=0}^c f_n(k) > \sum_{k=0}^\infty f(k) - 2ε$ for any $n \in \nn_{\ge m}$.
Together the above two facts imply that $\sum_{k=0}^c f_n(k) \to \sum_{k=0}^\infty f(k)$ as $n \to \infty$.
Notes
It is instructive to see why each of the conditions for MCT is important. It is also related to corresponding theorems in measure theory such as MCT for sets or MCT for functions.
A: When x tends to infinity, the first term containing x/x is obviously equal to one. Rest all terms are of the form, "something between zero and 1 raised to power infinity". So all other terms are going to be zero. 
Therefore answer should be just '1'.
