Applying squeeze/sandwich theorem to find $\lim_\limits{n\to\infty} \frac{1}{n^2} \sum_\limits{k=n}^{5n} k$ $$\lim_{n\to\infty}\Bigl( \frac{n}{n^2}+\frac{n+1}{n^2}+\cdots+\frac{5n}{n^2}\Bigr)$$
I know how to directly compute it, but I am required to use squeeze theorem.
Let $S_n$ be the sequence concerned.
I tried $(4n+1)\frac{n}{n^2}<S_n<(4n+1)\frac{5n}{n^2}$, but it gives me $4<\lim S_n<20$. What should I do to get appropriate bounds so that squeeze theorem can be applied?
 A: Compare the sum $A_n=\sum^{5n}_{k=n}k$ and the area of the trapezoidal $P_n$ made of by the positive side of the $x$-axis, the identity line $y=x$, and the vertical line $x=n$ and $x=5n$
splitting the  segment $[n,5n]$ in $4n$ pieces of length $1$.

*

*For each $n\leq k < 5n$,  we can construct the rectangles with base $[k,k+1]$, and height $k+1$. The sums of the areas of such rectangles is
$\sum^{5n}_{k=n+1}k$ is larger than the are of the $P_n$, which is $4n\frac{n+4n}{2}=12n^2$. This gives
$$ 12n^2 < A_n$$

*Similarly, for each $n\leq k < 5n$,  we can construct the rectangles with base $[k,k+1]$, and height $k$. The sums of the areas of such rectangles is
$\sum^{5n-1}_{k=n}k$ is smaller than the are of the $P_n$, which is $\frac{4n(n+4n)}{2}=12n^2$. This gives
$$ A_n-(5n-1)<12n^2$$
Putting things together
$$ 12 <\frac{1}{n^2}A_n <12 +\frac{5n-1}{n^2}$$
A: $$S=\lim_{n\to\infty}\Bigl( \frac{n}{n^2}+\frac{n+1}{n^2}+\cdots+\frac{5n}{n^2}\Bigr)$$
Note that the average of $n$ and $5n$ is $3n$
$$\lim_{n\to\infty;h\to 0^+}\Bigl( \frac{3n-h}{n^2}+\frac{3n-h}{n^2}+\cdots+\frac{3n-h}{n^2}\Bigr) < S < \lim_{n\to\infty;h\to0^+}\Bigl( \frac{3n+h}{n^2}+\frac{3n+h}{n^2}+\cdots+\frac{3n+h}{n^2}\Bigr)$$
$$\lim_{n\to\infty;h\to 0^+} 12-\frac{4h}{n} < S < \lim_{n\to\infty;h\to 0^+}12+\frac{4h}{n}$$
$$\implies \boxed{S=12}$$
Also note that a tight bound like this is required to evaluate the sum using squeeze theorem.
