Equivalent of $ u_{n}=\sum_{k=1}^n (-1)^k\sqrt{k}$ I'm trying to show that $$ u_{n}=\sum_{k=1}^n (-1)^k\sqrt{k}\sim_{n\rightarrow \infty} (-1)^n\frac{\sqrt{n}}{2}$$ when $n\rightarrow\infty$
How can I first show that $$u_{2n}\sim_{n\rightarrow \infty} \frac{\sqrt{2n}}{2}$$ and then deduce the equivalent of $u_{n}$?
 A: We have estimations
$$
u_{2n}=\sum\limits_{k=1}^{2n}(-1)^k\sqrt{k}=
\sum\limits_{k=1}^{n}(\sqrt{2k}-\sqrt{2k-1})=
\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k-1}}\leq
$$
$$
\sum\limits_{k=1}^{n}\frac{1}{2\sqrt{2k-1}}\leq
\int\limits_{1}^n\frac{dx}{2\sqrt{2(x+1)-1}}=\frac{\sqrt{2n+1}-\sqrt{3}}{2}
$$
and
$$
u_{2n}=\sum\limits_{k=1}^{2n}(-1)^k\sqrt{k}=
\sum\limits_{k=1}^{n}(\sqrt{2k}-\sqrt{2k-1})=
\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k-1}}\geq
$$
$$
\sum\limits_{k=1}^{n}\frac{1}{2\sqrt{2k}}\geq
\int\limits_{1}^n\frac{dx}{2\sqrt{2x}}=\frac{\sqrt{2n}-\sqrt{2}}{2}
$$
hence $u_{2n}\sim\frac{1}{2}\sqrt{2n}$ while $n\to\infty$.
A: Pair consecutive terms together: 
$$u_{2n} = \sum_{k=1}^{2n} (-1)^k \sqrt{k} = \sum_{k=1}^n \left( \sqrt{2k}-\sqrt{2k-1}\right) .$$
Since $\displaystyle \sqrt{1 - \frac{1}{2k}} = 1 - \frac{1}{4k} + \mathcal{O}(k^{-2})$ and $\displaystyle \sum_{k=1}^n k^p = \frac{n^{p+1}}{p+1} + \frac{n^p}{2} + \mathcal{O}(n^{p-1})$ we get 
$$u_{2n}= \sum_{k=1}^n \left(\frac{\sqrt{2}}{4\sqrt{k}} + \mathcal{O}(k^{-3/2})\right)= \frac{\sqrt{2n}}{2} + \mathcal{O}(1).$$
Now $u_{2n+1} = u_{2n} - \sqrt{2n+1}$ so the result that actually holds is $\displaystyle u_n \sim (-1)^n \frac{\sqrt{n}}{2}.$
