My teacher says that the answer to this questions is (a) and (d) but I'm getting (c) and (d). Please post a solution to this problem.
If $T_n = \displaystyle\sum_{r = 2n}^{(3n-1)}\left(\dfrac{r}{r^2+n^2}\right)$, $S_n = \displaystyle\sum_{r = 2n+1}^{3n}\left(\dfrac{r}{r^2+n^2}\right)$ then $\forall \ n \in \mathbb{N} :$
$(A) \ T_n > \dfrac{1}{2}\ln 2 \hspace{0.3 in} (B) \ S_n > \dfrac{1}{2}\ln 2 \hspace{0.3 in} (C) \ T_n < \dfrac{1}{2}\ln 2 \hspace{0.3 in} (D) \ S_n < \dfrac{1}{2}\ln 2$
On
This is off-topic but it is too long for a comment
Sooner or later, you will learn about the polygamma functions. So, writing $$\frac r {r^2+n^2}=\frac{1}{2 (r+i n)}+\frac{1}{2 (r-i n)}$$ you would find $$2T_n=-\psi ((2-i) n)-\psi ((2+i) n)+\psi ((3-i) n)+\psi ((3+i) n)$$ $$2S_n=-\psi ((2-i) n+1)-\psi ((2+i) n+1)+\psi ((3-i) n+1)+\psi ((3+i) n+1)$$ Using the asymptotics of the digamma function, you would find $$T_n=\frac 12\log(2)+\frac{1}{20 n}+\frac{1}{300 n^2}+O\left(\frac{1}{n^4}\right)$$ $$S_n=\frac 12\log(2)-\frac{1}{20 n}+\frac{1}{300 n^2}+O\left(\frac{1}{n^4}\right)$$
Hint: The function $f(x) = \dfrac{x}{x^2+1}$ is decreasing over $[2,3]$.
Hence, a left endpoint Riemann sum for $\displaystyle\int_{2}^{3}\dfrac{x}{x^2+1}\,dx$ will be larger than the integral, and a right endpoint Riemann sum for $\displaystyle\int_{2}^{3}\dfrac{x}{x^2+1}\,dx$ will be smaller than the integral.
Can you apply this to the problem by writing $T_n$ and $S_n$ in the form of a Riemann sum?