<s>[INST] Range of $S,T$ in $2$ sums 
If $\displaystyle S = \sum^{4n-1}_{k=3n}\bigg(\frac{k^2-7kn+13n^2}{n^3}\bigg)$ and $\displaystyle T= \sum^{4n}_{k=3n+1}\bigg(\frac{k^2-7kn+13n^2}{n^3}\bigg)$. then which one is/are
right  $\; (a)\displaystyle \; S<\frac{5}{6}\; (b)\; T<\frac{5}{6}\; (c)\; S>\frac{5}{6}\; (d)\; T>\frac{5}{6}$

Try: assuming for $$\lim_{n\rightarrow \infty}S = \lim_{n\rightarrow \infty}\sum^{4n-1}_{k=3n}\bigg(\frac{k^2-7kn+13n^2}{n^3}\bigg) =\lim_{n\rightarrow \infty}\frac{1}{n} \sum^{4n-1}_{k=3n}\bigg[\bigg(\frac{k}{n}\bigg)^2-7\bigg(\frac{k}{n}\bigg)+13\bigg]$$
conversion into definite integration  by substituting $\displaystyle \frac{k}{n}=x$ and $\displaystyle \frac{1}{n}=dx$
$$ = \int^{4}_{3}(x^2-7x+13)dx = \frac{5}{6}$$
could some help me how to estimate $S$ and $T$ , thanks
 [/INST] First,\begin{align*}
S - T &= \sum_{k = 3n}^{4n - 1} \frac{1}{n^3} (k^2 - 7kn + 13n^2) - \sum_{k = 3n + 1}^{4n} \frac{1}{n^3} (k^2 - 7kn + 13n^2)\\
&= \frac{1}{n^3} ((3n)^2 - 7 \cdot 3n \cdot n + 13n^2) - \frac{1}{n^3} ((4n)^2 - 7 \cdot 4n \cdot n + 13n^2)\\
&= \frac{1}{n} - \frac{1}{n} = 0.
\end{align*}
Now,\begin{align*}
S &= \sum_{k = 3n}^{4n - 1} \frac{1}{n^3} (k^2 - 7kn + 13n^2) = \frac{1}{n^3} \sum_{k = 3n}^{4n - 1} k^2 - \frac{7}{n^2} \sum_{k = 3n}^{4n - 1} k + 13\\
&\geqslant \frac{1}{n^3} \int_{3n}^{4n} x^2 \,\mathrm{d}x - \frac{7}{n^2} \sum_{k = 3n}^{4n - 1} k + 13\\
&= \frac{1}{n^3} \cdot \frac{1}{3} ((4n)^3 - (3n)^3) - \frac{7}{n^2} \cdot \frac{1}{2} (3n + (4n - 1)) n + 13\\
&= \frac{37}{3} - \frac{49n - 7}{2n} + 13 > \frac{37}{3} - \frac{49}{2} + 13 = \frac{5}{6}.
\end{align*}
Therefore,$$
S = T > \frac{5}{6}.
$$
</s>