how to prove this limit is zero? $$\lim_{n\to\infty}(\frac{\sqrt{1\times2}}{n^2+1}+\frac{\sqrt{2\times3}}{n^2+2}+\cdots+\frac{\sqrt{n\times(n+1)}}{n^2+n})=0$$
I tried to use $\sqrt{k(k+1)}\le\frac{2k+1}{2}$ or $k(k+1)=\sqrt{k}\times\sqrt{k}\times(k+1)\le\left(\frac{2\sqrt{k}+k+1}{3}\right)^3$ but nothing worked out. Can you show me some hint to deal with $\frac{\sqrt{k(k+1)}}{n^2+k}$ ? Thanks!
 A: If you enjoy harmonic numbers, starting with 
$$\sum \limits_{i=1}^{n}\dfrac{i}{n^2+i}  \lt \sum \limits_{i=1}^{n}\dfrac{\sqrt{i(i+1)}}{n^2+i}\lt \sum \limits_{i=1}^{n}\dfrac{i+1}{n^2+i}$$ you have
$$S_1=\sum \limits_{i=1}^{n}\dfrac{i}{n^2+i}=n+n^2\left(H_{n^2}- H_{n^2+n}\right)$$
$$S_2=\sum \limits_{i=1}^{n}\dfrac{i+1}{n^2+i}=n+(n^2-1)\left(H_{n^2}- H_{n^2+n}\right)$$ Now, using the asymptotics and continuing with Taylor series
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^3}\right)$$ for large values of $n$, you should have
$$S_1=\frac{1}{2}+\frac{1}{6 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
$$S_2=\frac{1}{2}+\frac{7}{6 n}-\frac{3}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
A: The limit is not $0$, it is $1/2$. Observe that
$$ \sum_{i=1}^{n} \frac{i}{n^2+i} < \sum_{i=1}^{n} \frac{\sqrt{i(i+1)}}{n^2+i} < \sum_{i=1}^{n} \frac{i+1}{n^2+i}$$
The limit of the left hand side is the same as the limit as the right hand.  By Squeeze theorem, the middle limit is also the same. So, it suffices to find the left hand limit.
$$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{i}{n^2+i} \geq \lim_{n \to \infty} \sum_{i=1}^{n} \frac{i}{n^2+n} = \lim_{n \to \infty} \frac{n(n+1)}{2n(n+1)} = \frac{1}{2}$$
This gives $1/2$ as a bound from below. We obtain $1/2$ as a bound above by observing $$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{i}{n^2+i} \leq  \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \frac{i}{n} = \int_0^1 x \ dx = \frac{1}{2}$$
The last step is a Riemann sum.
