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.
Best Answer
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.