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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

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

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