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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 $\forall \ n \in \mathbb{N} :$

$(A) \ T_n > \dfrac{1}{2}\ln 2 \hspace{0.3 in} (B) \ S_n > \dfrac{1}{2}\ln 2 \hspace{0.3 in} (C) \ T_n < \dfrac{1}{2}\ln 2 \hspace{0.3 in} (D) \ S_n < \dfrac{1}{2}\ln 2$

Please help me find the error in my solution :

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    Formatting matters. See [this page](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) for how to typeset your question properly. Also you should add some more info about why you get the answers you get. This will make it easier for people to spot your mistakes. – 2017-01-17
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    And please consider posting your solution (or atleast how did you proceed it with) so people here can help you better. – 2017-01-17
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    Since you did not give your solution, it is hard to see where is the problem. The only thing I can tell is that your teacher is right. – 2017-01-17
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    "Please help me find the error in my solution" **Which solution?** – 2017-01-17

2 Answers 2

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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?

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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 $$T_n=\frac 12\log(2)+\frac{1}{20 n}+\frac{1}{300 n^2}+O\left(\frac{1}{n^4}\right)$$ $$S_n=\frac 12\log(2)-\frac{1}{20 n}+\frac{1}{300 n^2}+O\left(\frac{1}{n^4}\right)$$

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    So according to your solution, Tn would be greater than 0.5ln(2) , right ? – 2017-01-17
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    @YashitGarg : Indeed. I did not want to do it with Riemann sums since you did not provide your steps but, menawhile, JimmyK4542 gave you the approach. – 2017-01-17