How do you solve this limit involving a sum?

Question

The limit which I'm having trouble with is this: $$\lim_{n\to\infty} \sum_{k=1}^{n}\frac{1}{n}\ln\left(\frac{k}{n}\right)$$ I tried substituting the $n$ at the top of the sum with $\infty$ and then tried manipulating it to figure out, but I kept getting $0$ as the answer. This is what I tried to evaluate: $$\sum_{k=1}^{\infty}\lim_{n\to\infty}\frac{1}{n}\ln\left(\frac{k}{n}\right)$$The limit actually goes to $-1$ though instead of $0 $, but I'm confused as to how to do it. I believe there is a way where you substitute the entire limit with an integral of something like this: $$\int_{0}^{1}\ln(x)dx$$ But I really want a method that doesn't use these pretty abstract methods to calculate, if there is a commonly known formula though that helps it simplify into this integral, that is OK.

Answer 1

You can't naively just replace the upper bound of the sum with $\infty$, because there's an $n$ inside of the sum as well. You need to be a little more clever.

In this case, by using logarithm properties, $$ \sum_{k=1}^n \frac 1 n \ln \frac k n = \frac 1 n \sum_{k=1}^n \ln \frac k n = \frac 1 n \ln \left( \prod_{k=1}^n \frac{k}{n} \right) = \frac 1 n \ln \left( \frac{n!}{n^n} \right) $$ The Stirling approximation tells us $$ n! \sim \sqrt{2\pi n} \frac{n^n}{e^n} $$ so as $n \to \infty$, $$ \frac 1 n \ln \left( \frac{n!}{n^n} \right) \sim \frac 1 n \ln \left( \frac{\sqrt{2\pi n} \frac{n^n}{e^n}}{n^n} \right) = \frac 1 n \ln \left( \sqrt{2\pi n} e^{-n} \right) = \frac{\ln \sqrt{2\pi} + \ln \sqrt n + \ln e^{-n}}{n} $$ Clearly, $$ \frac{\ln \sqrt{2\pi}}{n} \xrightarrow{n \to \infty} 0 \qquad \frac{\ln \sqrt n}{n} \xrightarrow{n \to \infty} 0 $$ leaving only $$ \frac{\ln e^{-n}}{n} = \frac{-n}{n} = -1 $$ for the limit. Hence, we conclude that the limit is $-1$.

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However, to be very frank, I would say that the Riemann sum approach is a lot less involved and straightforward (and not particularly abstract either). Since $$ \int_a^b f(x) \, \mathrm{d}x = \lim_{n \to \infty} \frac{b-a}{n} \sum_{i=1}^n f(\xi_i) $$ where $\xi_i$ lie in the $i$th induced subinterval given by the partition of $[a,b]$ at $(n+1)$-equally spaced points, it is easy to see that, for your case, $$\begin{align*} f(x) &= \ln(x) \\ \xi_k &= \frac k n, \text{ a right-endpoint evaluation rule} \\ b &= 1 \\ a &= 0 \end{align*}$$ and then compute $\int_0^1 \ln(x) \, \mathrm{d}x$, a trivial exercise with integration by parts.

Answer 2

As an alternative, avoiding Stirling approximation, by Stolz-Cesaro

$$\frac{\sum_{k=1}^{n+1}\ln\left(\frac{k}{n+1}\right)-\sum_{k=1}^{n}\ln\left(\frac{k}{n}\right)}{(n+1)-n}=\sum_{k=1}^{n}\ln\left(\frac{n}{n+1}\right)=\ln\left[\left(1-\frac{1}{n+1}\right)^n\right]\to\ln \left(\frac1e\right)=-1$$

Answer 3

As an alternative, to avoid Stirling approximation, using that

$$\sum_{k=1}^n \frac 1 n \ln \frac k n = \ln \left[\left( \frac{n!}{n^n} \right)^\frac1n\right]$$

since by ratio-root criterion

$$\frac{a_{n+1}}{a_n} \rightarrow L\implies a_n^{\frac{1}{n}} \rightarrow L$$

in this case by $a_n=\frac{n!}{n^n}$ we have

$$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=\frac1{\left(1+\frac1n\right)^n}\to \frac1e \implies \ln \left(a_n^{\frac{1}{n}}\right) \to -1$$