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.
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 $x&gt;m$, 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: $$a_{n,k}=\delta_{n,k}=\begin{cases}1&amp;n=k\\0&amp;n\neq k\end{cases}$$

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.
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'.