Definite Integration and summation of Series

Question

Hello,

I need help regarding above question. I solved for $S_n$ and $T_n$ as we solve. I got $S_n = \pi/(3\sqrt{3}) = T_n$. But the correct answers given in the Book are $A$ and $D$.

Please explain the reason behind the correct answers. I am really thankful to all the users who take effort to help me, thank you.

Answer 1

Note that

$$S_n=\frac1n \sum_{k=1}^n \frac{1}{1+k/n+(k/n)^2}$$

is the outer Riemann sum for the integral $\int_0^1 \frac{1}{1+x+x^2}\,dx$, while

$$T_n=\frac1n \sum_{k=0}^{n-1} \frac{1}{1+k/n+(k/n)^2}$$

is the inner Riemann sum for the same integral $\int_0^1 \frac{1}{1+x+x^2}$.

Since the summand and corresponding integrand are monotonically decreasing

$$T_n\ge \underbrace{\int_0^1 \frac{1}{1+x+x^2}}_{=\pi/(3\sqrt{3})}\,dx\ge S_n$$

Answer 2

We have $$S_n < \lim_{n \to \infty} S_{\infty} = \lim_{n \to \infty} \sum_{k=1}^{n} \frac {1}{n} \frac {1}{1+ \frac {k}{n} +(\frac {k}{n})^2 }$$ $$ = \int_{0}^{1} \frac {\mathrm {d}x}{1+x+x^2} = \frac {\pi}{3\sqrt {3}} $$

Now $$T_n > \frac{\pi}{3\sqrt {3}} \text { as } h\sum_{k=0}^{n-1} f (kh) > \int_{0}^{1} f (x) \mathrm {d}x > h\sum_{k=1}^{n} f (kh) $$

Hope it helps.