Finding bounds on $\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}$ [duplicate]

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.

• this came in KVPY 2021,I wrote the exam so i remember Feb 2 '21 at 5:45
• The integral gives a tighter lower bound than $0$ Feb 2 '21 at 6:05
• The "lower bound" is also $1$ (use $k<2n+1$). Feb 2 '21 at 8:06
• Feb 2 '21 at 8:23

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 $$\frac{859}{800}= 1.07375$$
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.