Evaluate $\lim_{n\ \rightarrow\ \infty\ }\ \left(\frac{\sqrt{1\times3}+\sqrt{2\times4}+\cdots\sqrt{n\times\left(n+2\right)}}{n}-\frac{n}{2}\right)$

Question

Evaluate $$\begin{align}\lim_{n\ \rightarrow\ \infty\ }\ \left(\frac{\sqrt{1\times3}+\sqrt{2\times4}+\cdots\sqrt{n\times\left(n+2\right)}}{n}-\frac{n}{2}\right)\end{align}$$
My Approach
Let $\begin{align}S=\sqrt{1\times3}+\sqrt{2\times4}+\cdots\sqrt{n\times(n+2)}\end{align}$
$\begin{align}1+2+\cdots+n<S<2+3+\cdots+(n+1)\end{align}$
$\begin{align}\frac{n(n+1)}{2}<S<\frac{n(n+3)}{2}\end{align}$
$\begin{align}\frac{1}{2}<\frac{S}{n}-\frac{n}{2}<\frac{3}{2}\end{align}$

But it is failed to apply squeezing principle.

Please I need some help with it, thanks.

Edit: RHS as above. About LHS, By GM $\geq$ HM, $\begin{align}S = \sum\limits_{k = 1}^n {\sqrt {k\left( {k + 2} \right)} } \ge \sum\limits_{k = 1}^n {\frac{{k\left( {k + 2} \right)}}{{\left( {k + 1} \right)}}} = \sum\limits_{k = 1}^n {\frac{{{{\left( {k + 1} \right)}^2} - 1}}{{\left( {k + 1} \right)}}} = \sum\limits_{k = 1}^n {\left( {k + 1} \right)} - \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} \end{align}$

Therefore
$\begin{align}\frac{3}{2} - \frac{{\sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} }}{n} < \frac{S}{n} - \frac{n}{2} < \frac{3}{2}\end{align}$

And hence the answer is $\frac{3}{2}$ by squeezing principle.

Answer 1

As suggested in the comments we can proceed by Riemann integral or as an alternative as follows

$$\frac{\sum_{i=1}^n \sqrt{i\cdot (i+2)}}{n} -\frac n 2 = \frac{2\sum_{i=1}^n \sqrt{i\cdot (i+2)}-n^2}{2n}$$

and by Stolz-Cesaro

$$\frac{2\sum_{i=1}^{n+1} \sqrt{i\cdot (i+2)}-2\sum_{i=1}^n \sqrt{i\cdot (i+2)}-(n+1)^2+n^2}{2(n+1)-2n}=$$$$=\frac{2\sqrt{(n+1)\cdot (n+3)}-2n-1}{2}$$

and we can finally take the limit.

Answer 2

To proceed by squeeze theorem we can use that for $0<x\le 2$

$$1+\frac x 2 -\frac{x^2}8 \le \sqrt{1+x}\le 1+\frac x2$$

then

$$\sum_{i=1}^n i+1-\frac 1{2i}\le \sum_{i=1}^n \sqrt{i\cdot (i+2)}=\sum_{i=1}^n i\sqrt{1+\frac 2i}\le \sum_{i=1}^n i+1$$

that is

$$\frac{n(n+1)}2+n -\frac12H_n \le \sum_{i=1}^n \sqrt{i\cdot (i+2)}\le \frac{n(n+1)}2+n$$

and then

$$ \frac 32-\frac12 \frac{H_n}n\le \frac{\sum_{i=1}^n \sqrt{i\cdot (i+2)}}{n} -\frac n 2 \le \frac 32$$

from which we can conclude.