# Find $\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} \ln\left(\frac{k}{n} + \epsilon_n\right)$ if $\epsilon_n>0$ and $\epsilon_n\to0$

Let $$\epsilon_{n}$$ be a sequence of positive reals with $$\lim\limits_{n \rightarrow \infty} \epsilon_{n}=0$$. Then find $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln\left(\frac{k}{n} + \epsilon_n\right)$$

My doubt here is that , can I introduce the limit within the summation? Then it can be easily found by converting the sum into an integral.

• I don't know what you mean by "introduce". The limit can't be moved inside the summation, as that leaves the $1/n$ dangling outside. But the limit certainly applies to the whole thing, including the summation. I would recommend trying to view the thing as a Riemann sum. – Gerry Myerson May 4 '17 at 7:21

Via mean value theorem we can see that $$\log\left(\frac{k} {n} +\epsilon_{n} \right) = \log\frac{k} {n} +\frac{n} {k} \epsilon_{n} +o(\epsilon_{n})$$ and hence the sum in question is equal to $$\frac{1}{n}\sum_{k=1}^{n}\log\frac{k}{n}+\epsilon_{n}\sum_{k=1}^{n}\frac{1}{k}+o(\epsilon_{n})$$ First sum tends to $\int_{0}^{1}\log x\, dx=-1$, second term tends to $\lim_{n\to\infty} \epsilon_{n} \log n$ so that the desired limit requires more information on the limiting behavior of $\epsilon_{n}$. For example if $\epsilon_{n} =1/n$ then the desired limit is $-1$ and if $\epsilon_{n} =1/\log n$ then the desired limit is $0$.

• I don't think that analyizng the second term is enough. Even if the second term is divergent, the third term ($-\frac12 n \epsilon_n^2 \sum_{k=1}\frac{1}{k^2}$) or any of the subsequent terms may also be divergent, and the divergences may cancel out, making the whole sum convergent in the end. Similarily if the second term is convergent, it doesn't determine the convergence of the subsequent terms. $\epsilon_n=\frac{1}{\log n}$ is such a case; the second term is covergent to $-1$, but the third $\sim \frac{n}{(\log n)^2}$ is divergent, as are all the subsequent ones. – Adam Latosiński May 29 at 7:30
• Also, you can expand $\log(1+\frac{n\epsilon_n}{k} ) =\frac{n\epsilon_n}{k} + O(\epsilon_n^2)$ only if $\frac{n\epsilon_n}{k} < 1$. If $n\epsilon_n > 1$, then for a given $n$ you can only make such expansion for some $k$, but not for all. – Adam Latosiński May 29 at 8:08

For any $$\delta>0$$, for $$n$$ big enough $$\frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n}\right) \le \frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n} + \epsilon_n\right) \le \frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n} + \delta\right)$$ so $$\lim_{n\rightarrow \infty}\frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n}\right) \le \lim_{n\rightarrow \infty}\frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n} + \epsilon_n\right) \le \lim_{n\rightarrow \infty}\frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n} + \delta\right)$$ $$\int_0^1 \log(x) dx \le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n} + \epsilon_n\right) \le \int_0^1 \log(x+\delta) dx$$ $$-1 \le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n} + \epsilon_n\right) \le (1+\delta)\log(1+\delta) - \delta\log\delta -1$$ $$\delta$$ is arbitrary, and $$\lim_{\delta\rightarrow 0}\big((1+\delta)\log(1+\delta) - \delta\log\delta -1 \big) = -1$$, so $$\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n} + \epsilon_n\right) = -1$$