<s>[INST] Comparison of $S_{n}$ and $T_{n}$, where $S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}$ and $T_{n} = \sum_{k=1}^{n-1}\frac{n}{n^2+kn+k^2}$ Let $\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}$ and $\displaystyle T_{n} = \sum_{k=1}^{n-1}\frac{n}{n^2+kn+k^2}$ for $n=1,2,3,\dots$
Then which of the following options are Right.
Options::  $\displaystyle (a)\; S_{n}<\frac{\pi}{3\sqrt{3}}\;\;\;\;\;\;\;\;(b)\;S_{n}>\frac{\pi}{3\sqrt{3}}\;\;\;\;\;\;\;\;(c)\; T_{n}<\frac{\pi}{3\sqrt{3}}\;\;\;\;\;\;\;\;(d)\; T_{n}<\frac{\pi}{3\sqrt{3}}\;\;\;\;\;\;\;\;$
$\bf{My\; Try}::$ Given $\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}<\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2} = \int_{0}^{1}\frac{1}{1+x+x^2}dx$
(By converting above infinite sum into Integral).
Now Let $\displaystyle I = \int_{0}^{1}\frac{1}{1+x+x^2}dx = \int_{0}^{1}\frac{1}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx = \frac{2}{\sqrt{3}}\cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}}$
So $\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}<\frac{\pi}{3\sqrt{3}}$
Now How can I solve for $\displaystyle T_{n} =\sum_{k=1}^{n-1}\frac{n}{n^2+kn+k^2}$
Help Required
Thanks
 [/INST] \begin{align*}
T_n&=\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2}-\frac{1}{n}\\
&<\lim_{n\rightarrow \infty}\left(\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2} -\frac{1}{n}\right)= \int_{0}^{1}\frac{1}{1+x+x^2}dx+0=\frac{\pi}{3\sqrt{3}}
\end{align*}
OR
\begin{align*}
T_n&=\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}-\frac{1}{3n}\\
&<\lim_{n\rightarrow \infty}\left(\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2} -\frac{1}{3n}\right)= \int_{0}^{1}\frac{1}{1+x+x^2}dx+0=\frac{\pi}{3\sqrt{3}}
\end{align*}
</s>