Limit of the expression I want to find following limit 
$$\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{\infty}|\ln n-\ln k|\left(1-\frac{1}{n}\right)^k=?$$
To use computer programs is also allowed. Thanks for your helps...
 A: For a geometric distribution with parameter $p$ we have:
$$\mathbb{P}[X=k]=p(1-p)^{k-1}$$
for $k=1,2,3,\ldots$. If the parameter is $p=\frac{1}{n}$ and $n\to +\infty$ the limit distribution is an exponential distribution, so we have:
$$\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{\infty}|\ln n-\ln k\,|\left(1-\frac{1}{n}\right)^k=\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{+\infty}\left|\log\frac{k}{n}\right|e^{-k/n},\tag{1}$$
but the last sum is a Riemann sum, converging to:
$$ \int_{0}^{+\infty}\left|\log x\right|e^{-x}\,dx = \gamma-2\,\text{Ei}(-1) = \color{red}{1.0159835336925734\ldots},\tag{2}$$ 
where $\gamma$ is the Euler-Mascheroni constant and $\text{Ei}$ is the exponential integral. 
Such a limit has the following fast-converging series representation:
$$ -\gamma-2\sum_{k=1}^{+\infty}\frac{(-1)^k}{k\cdot k!}.\tag{3}$$
A: To get to the improper integral directly: We have
$$S_n = \frac{1}{n}\sum_{k=1}^{\infty}|\ln (k/n)|[(1-1/n)^n]^{k/n}.$$
Now $(1-1/n)^n$ increases to $1/e$ as $n$ increases to $\infty.$ Let $0<a<1/e.$ Then for large $n$ we have
$$\frac{1}{n}\sum_{k=1}^{\infty}|\ln (k/n)|a^{k/n} \le S_n \le \frac{1}{n}\sum_{k=1}^{\infty}|\ln (k/n)|e^{-k/n}.$$
Let $n\to \infty$ to get 
$$\int_0^\infty |\ln x|a^{x}\,dx \le \liminf S_n \le \limsup S_n \le \int_0^\infty |\ln x|e^{-x}\,dx.$$
This is true for any $0<a<1/e.$ Now let $a$ increase to $1/e$ and use the dominated convergence theorem to see the desired limit of $S_n$ is the integral on the right above.
