Let S_n = sum_{k=1}^n frac{n}{n^2+kn+k^2} and | vedclass

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(A,D) We can rewrite the sums as Riemann sums: $S_n = \sum_{k=1}^n \frac{1}{n} \cdot \frac{1}{1 + (k/n) + (k/n)^2}$ $T_n = \sum_{k=0}^{n-1} \frac{1}{n

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$S_n < \frac{\pi}{3\sqrt{3}}$

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(A,D) We can rewrite the sums as Riemann sums: $S_n = \sum_{k=1}^n \frac{1}{n} \cdot \frac{1}{1 + (k/n) + (k/n)^2}$ $T_n = \sum_{k=0}^{n-1} \frac{1}{n} \cdot \frac{1}{1 + (k/n) + (k/n)^2}$ Let $f(x) = \frac{1}{1+x+x^2}$. Since $f(x)$ is a strictly decreasing function on $[0, 1]$,the right Riemann sum $S_n$ is an underestimate of the integral,and the left Riemann sum $T_n$ is an overestimate. Thus,$S_n Evaluating the integral: $\int_0^1 \frac{dx}{(x+1/2)^2 + 3/4} = \left[ \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2x+1}{\sqrt{3}} ight) ight]_0^1 = \frac{2}{\sqrt{3}} (	an^{-1}(\sqrt{3}) - 	an^{-1}(1/\sqrt{3})) = \frac{2}{\sqrt{3}} (\pi/3 - \pi/6) = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}}$. Therefore,$S_n  \frac{\pi}{3\sqrt{3}}$.

Let $S_n = \sum_{k=1}^n \frac{n}{n^2+kn+k^2}$ and $T_n = \sum_{k=0}^{n-1} \frac{n}{n^2+kn+k^2}$ for $n=1, 2, 3, \ldots$. Then,

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