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We have to show that $$ \lim_{n\to\infty} \left( \dfrac 1 {\sqrt{n^2 +1}} +\dfrac 1 {\sqrt{n^2 +2}} + \dfrac 1 {\sqrt{n^2 +3}} + \ldots + \dfrac 1 {\sqrt{n^2 +n}} \right) = 1$$

Now, I thought of doing it as a definite integral, however the terms don't translate into a function of $\left(\dfrac r n\right)$. I tried searching wolfram-alpha, but it used Hurwitz-zeta function, of which I have no clue. (This is an assignment problem)

Git Gud
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Parth Thakkar
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1 Answers1

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Hint: For all $n\in \mathbb N$ it holds that

$$ \dfrac n {\sqrt{n^2 +n}} \leq \dfrac 1 {\sqrt{n^2 +1}} +\dfrac 1 {\sqrt{n^2 +2}} + \dfrac 1 {\sqrt{n^2 +3}} + \ldots + \dfrac 1 {\sqrt{n^2 +n}}\leq \dfrac n {\sqrt{n^2 +1}}.$$

Git Gud
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