I was asked this series
$$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n \log \left( 1+\frac{2i}{n} \right) $$
I tried to use Riemann sum
$$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n \log \left( 1+\frac{2i}{n} \right) = \lim_{n\to\infty} \int_{1}^{n} \log(1+2x) \, \mathrm{d}x $$
But the answer is divergent. What is wrong here?
When you write an integral $\int_a^b f(x)\,\mathrm dx$ into a Riemann sum, what you essentially do is: you partition the interval $[a,b]$ into $n$ segments, each of length $h:=(b-a)/n$ and you sum up the area of the rectangles formed by the partition that converges to the integral when the partition is made finer by letting $n\to\infty$. Thus, $$\int_a^b f(x)\,\mathrm dx=\lim_{n\to\infty}\sum_{i=0}^{n-1}hf(a+ih)=\lim_{n\to\infty}\sum_{i=1}^n hf(a+ih)$$ where $h:=(b-a)/n$ is the width of each partition segment. The sum from $0...(n-1)$ is the lower Riemann sum and the one from $1...n$ is the upper Riemann sum.
Now, a question to you: what should $a,b$ be for your problem so that $h:=\frac{b-a}n$ equals $1/n$ and the term $f(a+ih)$ in the summation equals $f(i/n)$? And after that, what choice of $f(x)$ fits such that $f(i/n)$ equals the summation term $\log\left(1+2\cdot\frac in\right)$ in your problem?