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.
 A: 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 $0$.
A: 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
$$ \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n} + \epsilon_n\right) = -1$$
