Two series and corresponding Riemann Sums

Question

Let $$S_{n}=\sum_{r=n}^{3n-1}\frac{n}{r^2-4rn}$$ and $$T_{n}=\sum_{r=n+1}^{3n}\frac{n}{r^2-4rn}$$

To prove that $$S_{n}<ln\frac{1}{\sqrt{3}}$$ and $$T_{n}<ln\frac{1}{\sqrt{3}}$$

I am having difficulty converting the sums to Riemann Integrals

Answer 1

First observe $S_n = T_n$: $$\begin{align} S_n &= \sum_{r=n}^{3n-1} \frac{n}{r (r-4n)}\\ &= \sum_{t={n+1}}^{3n} \frac{n}{(4n-t)(-t)} \\ &= \sum_{r=n+1}^{3n} \frac{n}{r (r-4n)}\\ &= T_n \end{align}$$ where substitution $t = 4n-r$ was used, then dummy variable changed. So, it suffices to show $\frac{S_n + T_n}{2}< -\frac{1}{2} \ln 3$.

We know $$\frac{S_n + T_n}{2} = \frac{1}{2n}\left[f(1) + 2\sum_{r=n+1}^{3n-1} f\left({r\over n}\right) + f(3)\right]$$ where $f(x) = {1 \over x^2 - 4x}$, which you can recognise as the trapezium rule. But $f$ is convex, so the approximation is always less than the integral, hence $$\frac{S_n + T_n}{2} < \int_1^3 \frac{1}{x^2 - 4x} dx=-\frac{1}{2} \ln 3$$ as required.

Answer 2

Both sums will give the same integral when you take the limit \begin{eqnarray*} \sum_{r=n}^{3n} \frac{n}{r^2-4nr} = \frac{1}{n} \sum_{r=n}^{3n} \frac{1}{(r/n)^2-4r/n} \rightarrow \int_{1}^{3} \frac{dx}{x^2-4x}. \end{eqnarray*} Now partial fractions \begin{eqnarray*} \frac{1}{x(x-4)} =\frac{1}{4} \left( \frac{1}{x-4} - \frac{1}{x} \right). \end{eqnarray*} The integral is \begin{eqnarray*} \int_{1}^{3} \frac{dx}{x^2-4x} =\frac{1}{4} \left[ \ln(x-4) - \ln(x) \right]_1^3 = -\frac{1}{2} \ln3. \end{eqnarray*}