I have to compute the limit of this sequence ${\textstyle\sum_{k=1}^n}\frac1{\sqrt{n^2+k}}$ as $n\rightarrow\infty$. First I was thinking about some Riemann sum and and forced the $n^{2}$ outside the square root but the function was not so pleasant.
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1Does this series converge? – Sean Nemetz Apr 09 '18 at 15:53
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@SeanNemetz Yes. – C. Cristi Apr 09 '18 at 15:54
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1@C.Cristi please refrain from calling $\sum_{k=1}^n \frac1{\sqrt{n^2+k}}$ a series. This is merely a sequence. A series would be $\sum_{k=1}^n a_k$ where $a_k$ does not depend on $n$. – Gabriel Romon Apr 09 '18 at 16:02
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@GabrielRomon Yeah, you're right. My bad.. – C. Cristi Apr 09 '18 at 16:03
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How about squeezing ?
$$\frac{n}{\sqrt{n^2+n}}\leq \sum_{k=1}^n\frac1{\sqrt{n^2+k}}\leq \frac{n}{\sqrt{n^2+1}}$$
The outer terms both go to $1$.
Gabriel Romon
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Hmm, yeah. Can you tell me more about squeezing theorem and how to use it when you're dealing with this kind of fractions? – C. Cristi Apr 09 '18 at 16:02
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@C.Cristi A sum of $n$ terms is less than $n\cdot \text{biggest summand}$ and greater than $n\cdot \text{smallest summand}$. This is a general principle that doesn't necessarily yield sharp bounds. For your problem though, it's sufficient. – Gabriel Romon Apr 09 '18 at 16:05