Limit of $u_n = \sum_{k=1}^{n}\frac{1}{\sqrt{n^2+2k}}$

I want to find the limit of this sequence if it exists :

$$u_n = \sum_{k=1}^{n}\frac{1}{\sqrt{n^2+2k}}$$

My attempt is to first remark that for $$k\in\{1,...,n\}$$ :

$$\frac{1}{\sqrt{n^2+2n}}\leq\frac{1}{\sqrt{n^2+2k}}\leq\frac{1}{\sqrt{n^2+2}}\implies\frac{n}{\sqrt{n^2+2n}}\leq\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+2k}}\leq\frac{n}{\sqrt{n^2+2}}$$ Which leads to : $$\frac{1}{\sqrt{1+\frac{2}{n}}}\leq\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+2k}}\leq\frac{1}{\sqrt{1+\frac{2}{n^2}}}\implies\lim\limits_{n\to\infty}u_n=\lim\limits_{n\to\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+2k}} = 1$$

The last implication follows from the fact that the square root function is continuous at $$x = 1$$ and the squeeze theorem.

I would like to know first if my attempt is correct and if you have other ideas to show this that can be enlightening !

Thank you a lot

• You forgot to type limit after $\implies$ but otherwise, your proof is fine. Jan 21, 2023 at 9:26
• Great, thank you for your comment ! Jan 21, 2023 at 9:35
• Just for you rcuriosity : If $n$ is large (say $n=5$) $$u_n=1-\frac{1}{2 n}+\frac{1}{8 n^3}+O\left(\frac{1}{n^4}\right)$$ Jan 21, 2023 at 9:49
• I would use $${1\over n+1}\le {1\over \sqrt{n^2+2k}}\le {1\over n}$$ Jan 21, 2023 at 10:24