The limit $n \rightarrow \infty$ of the standard deviation of $x_k= \ln k, k=1,2,3,..,n.$

Question

The standard deviation for a sequence $x_k$ is defined as $$S_n=\sqrt{\sum_{k=1}^{n} \frac{{x_k}^2}{n}-\left(\sum_{k=1}^{n} \frac{x_k}{n}\right)^2}$$ By numerics the asymptotic value of $S_n$ for $x_k=\ln k$ turns out to be: $S_n \sim 1.$ Now the question is: Can one show that $\lim_{n\rightarrow \infty} S_n =1$?

Answer 1

Replacing $x_k$ with $x_k-a$ (where $a$ doesn't depend on $k$) doesn't change $S_n$. Thus, $$S_n^2=\frac{1}{n}\sum_{k=1}^{n}\ln^2\frac{k}{n}-\left(\frac{1}{n}\sum_{k=1}^{n}\ln\frac{k}{n}\right)^2\underset{n\to\infty}{\longrightarrow}\int_0^1\ln^2x\,dx-\left(\int_0^1\ln x\,dx\right)^2=1.$$

Answer 2

A little longer way We have $$\lim_{n\rightarrow \infty}{S_n}^2=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{\ln^2 k}{n}- \lim_{n \rightarrow} \left( \sum_{k=1}^{n} \frac{\ln k}{n}\right)^2=L_1-L_2.......(1)$$ We can write $$L_1=\lim_{n\rightarrow n}\sum_{k=1}^{n} \frac{{\ln^2 k}}{n}=\lim_{n\rightarrow n}\sum_{k=1}^{n} \frac{{(\ln (k/n)+\ln n)^2}}{n}......(2)$$ $$\Rightarrow L_1=\lim_{n\rightarrow n}\sum_{k=1}^{n} \frac{{\ln^2 (k/n)}}{n}+\lim_{n\rightarrow \infty} \ln^2 n + \lim_{n\rightarrow \infty} \left( 2 \ln n \sum_{k=1}^{n} \frac{\ln (k/n)}{n} \right).......(3) $$ Now take up $$L_2=\left (\lim_{n\rightarrow \infty} \sum_{k=1}^{n} \frac{(\ln(k/n) +\ln n)}{n} \right)^2=\left(\lim_{n \rightarrow n} \sum_{k=1}^{n} \frac{\ln (k/n)}{n} \right)^2 +\lim_{n\rightarrow \infty} \ln^2 n + \lim_{n\rightarrow \infty} \left( 2 \ln n \sum_{k=1}^{n} \frac{\ln (k/n)}{n} \right)...(4)$$ From (3) and (4), we get $$L=\lim_{n\rightarrow n}\sum_{k=1}^{n} \frac{{\ln^2 (k/n)}}{n}-\left (\lim_{n\rightarrow \infty} \sum_{k=1}^{n} \frac{\ln(k/n)}{n}\right)^2.$$ Chnging limit of sum to integral, we wet $$L=\int_{0}^{1} \ln^2 x~dx-\left(\int _{0}^{1} \ln x~dx \right)^2.= \int_{0}^{\infty} t^2 e^{-t} dt - \left( \int_{0}^{\infty} t e^{-t} dt \right)^2=2-1=1.$$