Evaluate the sum $\sum^{\infty}_{n=2}{\frac{1}{n^3-n}}$ using Riemann Sum My lecturer has told us to calculate the sum of
$$\sum^{\infty}_{n=2}{\frac{1}{n^3-n}}$$
by using partial fractions and deal with telescoping series.
I have a thought of using Riemann sum to compute the value. With the width $\frac{1}{n}$, and range in between $2$ and $\infty$, I have to find a function $f(x)$ such that in every point of $x^*=2+\frac{i}{n}$, the area computed is $\frac{1}{n} \cdot f(x^*)$. 
I have a trouble finding such function, any clue that help in solving such problem?
 A: The method of Riemann sums is not suitable for finding this particular sum. A Riemann sum is a limit on the form 
$$\lim_{n\to\infty} \sum_{k=1}^n f(c_k)(x_k - x_{k-1})$$
where $a = x_0 < \ldots < x_n = b$ is some increasing (and eventually tightly spaced) sequence in an interval $[a,b]$ and where $c_k \in (x_{k-1},x_k)$. Your sum is not on this form and can very likely not be put on this form. 
Also note that Riemann sums are over finite intervals $[a,b]$, you can't have it over $[0,\infty)$ (other than as a limit $R\to\infty$ as a Riemann sum over $[0,R]$).
Thus I would just discard that method for this sum and move on. If you want to practice another method then you could for example try a power-series method. Motivated by the fact that $\frac{1}{n^3-n} = \frac{1}{(n+1)n(n-1)}$ we can define $f(x) = \sum_{k=2}^\infty \frac{x^{k+1}}{k^3-k}$ and by differentiation we simplify the summand and we end up with $f'''(x) = \sum_{k=2}^\infty x^{k-2} = \frac{1}{1-x}$. You can now integrate up three times (and carefully setting the integration constants each time) to get $f(x) = \frac{1}{4} \left(3 x^2-2 x+(4 x-2x^2-2) \log (1-x)\right)$ and then taking $x\to 1$ to get the sum you are after. In this case this method leads to a lot of calculation compared to telescoping so it's not the most efficient method, but it's useful to know about.
