Comparison: $S_{1} = \sum\nolimits_{r=n}^{3n-1}{\frac{n}{r^2-4rn} } $; $S_{2} = \sum\nolimits_{r=n+1}^{3n}{\frac{n}{r^2-4rn} }$ ; $-\frac{1}{2}\ln{3}$

Question

$$S_{1} = \sum\limits_{r=n}^{3n-1}{\frac{n}{r^2-4rn}} $$

$$S_{2} = \sum\limits_{r=n+1}^{3n}{\frac{n}{r^2-4rn}} $$

Compare value of $S_{1}$, $S_{2}$, $-\frac{1}{2}\ln{3}$.

I was trying and i figured out $S_{1} = S_{2}$.
but can't relate it with $-\frac{1}{2}\ln{3}$ which is, in fact, it's limiting value when n tends to infinity.

Answer 1

$$S(a,b,n) = \sum_{r=a}^b{\frac{n}{r^2-4rn}}=\frac 14 \sum_{r=a}^b\left(\frac{1}{r-4n} -\frac{1}r\right)$$

Using harmonic numbers $$S(a,b,n) =\frac 14\left(H_{b-4 n}-H_{a-4 n-1}+H_{a-1}-H_b \right)$$ Now, using $a=n$ and $b=3n-1$ (as for $S_1$).

Using asymptotics, you would find $$S(n,3n-1,n)=\frac \pi 4 (\cot(n \pi)-\cot(3n \pi))-\frac {\log(3)}2-\frac{1}{27 n^2}+\frac{1}{243 n^4}+O\left(\frac{1}{n^6}\right)$$ But, since $n$ is supposed to be an integer (in any manner $3n\pi=n\pi+2n\pi)$ the first term is $0$ and you are left with $$S(n,3n-1,n)=-\frac {\log(3)}2-\frac{1}{27 n^2}+\frac{1}{243 n^4}+O\left(\frac{1}{n^6}\right)$$

Answer 2

$\mathcal{Hint}$

Consider the function $f(x)=\frac{1}{x^2-4x}$ in the interval $[1,3]$.

Try to compare $\frac{S_1+S_2}{2}$(as sum of areas of trapezoids) with the definite integral $\int_1^3f(x)dx$.