Finding bounds on $\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}$ I am supposed to approximate the value of $\lim_{n\to \infty}\sum_{k=0}^{n}1/\sqrt{n^2+k}$. If it were $k^2$, then it could be rewritten as an integral, which would come out to be $\ln(\sqrt{2}+1)$.
$$\lim_{n\to \infty}\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}\lt \lim_{n\to \infty}\frac{n+1}{n}=1$$
So the limiting sum should lie in the interval $(0,1)$. Is this correct? Thanks.
 A: If you use generalized harmonic numbers
$$S_n=\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}=H_{n^2+n}^{\left(\frac{1}{2}\right)}-H_{n^2-1}^{\left(\frac{1}{2}\right)}$$ Now, using asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2\sqrt{p}}+O\left(\frac{1}{p^{3/2}}\right)$$ apply it twice and continue with Taylor series to obtain
$$S_n=1+\frac{3}{4 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$
For example, the "exact" value of $S_{10}$ is $\sim 1.07386$ while the above truncated series gives $\frac{859}{800}= 1.07375$
A: Hint
$$\dfrac{1}{\sqrt{n^2 + n}} \le \dfrac{1}{\sqrt{n^2 + k}}\le \dfrac{1}{\sqrt{n^2}}$$
for $0 \le k \le n$. Sum both sides from $0$ to $n$ and see that the sums are equal, in the limit.

The limit will turn out to be $1$. Your assertion that the limiting sum is in $(0, 1)$ is incorrect. Note that even if you have $a_n < b_n$ for convergent sequences, all you can conclude is that $\lim a_n \le \lim b_n$. That is, the strict inequality cannot be passed on to the limit.
