# Computing the limit of the sequence $\lim_{n\rightarrow\infty}\overset n{\underset{k=1}{\sum(}}\sqrt{1+\frac k{n^2}}-1)$ and another one [closed]

I have to compute the limit of this sequence $\lim\limits_{n\to\infty}$ $\overset n{\underset{k=1}{\sum(}}\sqrt{1+\frac k{n^2}}-1)$. I tried to apply it's conjugate and then I go to this point: $\lim\limits_{n\rightarrow\infty}\overset n{\underset{k=1}{\sum(}}\frac k{n\sqrt{n^2+k}+n})$ and then tried to apply squeezing theorem but it gets me to a wrong answer...

The other sequence is: $\lim\limits_{n\rightarrow\infty}\frac1{3n+1}\sum\limits_{k=1}^n\cos\frac\pi{2n+k}$, for which I have no idea how to start it.

If you answer me can you provide me with some information about Limits of sequences and how to deal with them? Especially the ones with sums of square roots and trig sums.

## closed as off-topic by Did, José Carlos Santos, Tom-Tom, Holo, user223391 Apr 13 '18 at 0:55

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## 2 Answers

Using $$1+\frac{x}{2}>\sqrt{1+x}>1+\frac{x}{2}-x^2$$ for $0<x<1$

For any $x\in(0,1)$ we have $\sqrt{1+x}-1=\frac{x}{2}+O(x^2)$, hence $$\sum_{k=1}^{n}\sqrt{1+\frac{k}{n^2}}-1 =\frac{1}{2n^2}\sum_{k=1}^{n}k+O\left(\frac{1}{n^4}\sum_{k=1}^{n}k^2\right)=\color{red}{\frac{1}{4}}+O\left(\frac{1}{n}\right)$$ as $n\to +\infty$.

• What do you mean through $O(x^{2})$ and $O(\frac {1}{n})$? – C. Cristi Apr 10 '18 at 13:01
• @C.Cristi: en.wikipedia.org/wiki/Big_O_notation – Jack D'Aurizio Apr 10 '18 at 13:04
• can you be a little more specific in your answer? – C. Cristi Apr 10 '18 at 13:10
• @C.Cristi: read $a=O(b)$ as $|a|\leq K |b|$ for some constant $K$ (whose value is not really relevant). – Jack D'Aurizio Apr 10 '18 at 13:27