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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.

  • 1
    $T_n-S_n=\frac{1}{2n} >0 $so $T_n>S_n$. but since the difference vansishs for infinite $n$ the limit of both sequences is the same $l=\lim{n\rightarrow\infty}(\frac1n\sum_{k\geq 0}s_{k/n})=\int_0^1s(x)dx$ – 2017-01-27

2 Answers 2

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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

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

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    nice work $(+1)$ – 2017-01-27
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    I am a beginner. Riemann sum is not in the syllabus. Question are framed in such a way which can be solved using DEFINITE INTEGRAL AS A LIMIT OF SUM and LIMIT AS DEFINITE INTEGRAL. I want to know if there is any logic which a beginner can apply? – 2017-01-27
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    The definite integral as the limit of a sum is a Riemann sum (i.e,. an integral). So, you actually have the right tool. – 2017-01-27
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    @tired Thank you. For some reason, I missed your comment. Apology. – 2017-01-28
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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.