I have to solve the limit $$\lim_{n\rightarrow\infty}\left(\frac1n\sum_{k=1}^n(\text{log}\:k)^2-\left(\frac1n\sum_{k=1}^n(\text{log}\:k)\right)^2\right)$$

I don't know how to proceed in this problem. However, the solution says that this limit is equal to the limit $$\lim_{n\rightarrow\infty}\left(\frac1n\sum_{k=1}^{n-1}\left(\text{log}\:\frac kn\right)^2-\left(\frac1n\sum_{k=1}^{n-1}\left(\text{log}\:\frac kn\right)\right)^2\right)$$

Which is equal to $$\int_0^1 (\text{log}\:x)^2dx-\left(\int_0^1 \text{log}\:x\;dx\right)^2$$

Can someone explain how these two equations were derived from the first expression? I am aware of expressing definite integral as limit of an infinite sum. But I don't see how the above conversions were made. I need the intermediate steps.



$$\log k = \log k - \log n + \log n = \log \frac{k}{n} + \log n,$$

so that

\begin{align}\left(\frac{1}{n}\sum_{k = 1}^n \log k\right)^2 &= \left(\frac{1}{n}\sum_{k = 1}^n \log\frac{k}{n} + \log n\right)^2\\ &= \left(\frac{1}{n}\sum_{k = 1}^n \log \frac{k}{n}\right)^2 + \frac{2\log n}{n}\sum_{k = 1}^n \log \frac{k}{n} + (\log n)^2 \tag{1} \end{align}


\begin{align}\frac{1}{n}\sum_{k = 1}^n (\log k)^2 &= \frac{1}{n}\sum_{k = 1}^n \left[\left(\log\frac{k}{n}\right)^2 + 2\log n\log\frac{k}{n} + (\log n)^2\right]\\ &=\frac{1}{n}\sum_{k = 1}^n \left(\log \frac{k}{n}\right)^2 + 2\frac{\log n}{n}\sum_{k = 1}^n \log \frac{k}{n} + (\log n)^2 \tag{2} \end{align}

Subtracting $(1)$ from $(2)$, we obtain

\begin{align} & \frac{1}{n}\sum_{k = 1}^n (\log k)^2 - \left(\frac{1}{n}\sum_{k = 1}^n \log k\right)^2 \\ &= \frac{1}{n}\sum_{k = 1}^n \left(\log \frac{k}{n}\right)^2 - \left(\frac{1}{n}\sum_{k = 1}^n \log \frac{k}{n}\right)^2\\ &= \frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2 - \left(\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n}\right)^2 \end{align}

The last step follows since $\log \frac{k}{n} = 0$ when $k = n$.


$$\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n}$$

is a sequence of Riemann sums for $\log x$ over $[0,1]$, and

$$\frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2$$

is a sequence of Riemann sums for $(\log x)^2$ over $[0,1]$, we have

$$\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n} \to \int_0^1 \log x\, dx$$


$$\frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2 \to \int_0^1 (\log x)^2\, dx,$$

whence the third step follows.

  • $\begingroup$ Can you please also include the last step of getting the integral? $\endgroup$
    – Tejas
    Mar 5 '15 at 3:46
  • $\begingroup$ @Tejas I have included the details for getting the last step. $\endgroup$
    – kobe
    Mar 5 '15 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.