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 $u_{n}$?

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 $n\to\infty$.
