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.

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