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 $u_{n}$?

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You can simply swap $2n$ with $n$ both in the formula and the index to deduce the equivalent. –  Pedro Tamaroff Jun 7 '12 at 18:04

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 $\displaystyle u_n \sim (-1)^n \frac{\sqrt{n}}{2}.$

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You cannot say $\sum_{k=1}^n O(k^{-3/2}) = O(n^{-1/2})$ ... it could be as bad as $O(1)$. –  GEdgar Jun 7 '12 at 18:43
@GEdgar Ahh yes, sorry. I'll fix that now. –  Ragib Zaman Jun 7 '12 at 18:46
@Ragib Zaman: I got it thank you very much for your answer! –  Chon Jun 7 '12 at 19:05

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 $n\to\infty$.

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Thank you very much Norbert! –  Chon Jun 7 '12 at 19:06
@Chone, you are welcome! –  Norbert Jun 7 '12 at 19:11

Norbert's answer is the simplest with the fewest prerequisites. That being the case, here is a nuke for an ant hill using the Euler-Maclaurin Sum Formula.

Note that \begin{align} \sum_{k=1}^{2n}(-1)^k\sqrt{k} &=2\sum_{k=1}^{n}\sqrt{2k}\;-\;\sum_{k=1}^{2n}\sqrt{k}\\ &=\sqrt{8}\sum_{k=1}^{n}\sqrt{k}\;-\;\sum_{k=1}^{2n}\sqrt{k}\tag{1} \end{align} The Euler-Maclaurin Sum Formula says that $$\sum_{k=1}^n\sqrt{k}=\frac23n^{3/2}+\frac12n^{1/2}+C+\frac{1}{24}n^{-1/2}+O\left(n^{-3/2}\right)\tag{2}$$ For $\mathrm{Re}(z)>-1$, $$\zeta(z)=\lim_{n\to\infty}\sum_{k=1}^nk^{-z}\;-\;\left(\frac{1}{1-z}n^{1-z}+\frac12n^{-z}\right)\tag{3}$$ Applying $(3)$ to $(2)$ yields $C=\zeta(-\frac12)=-\frac{1}{4\pi}\zeta(\frac32)\,\dot{=}-0.207886224977354566$.

Applying $(2)$ to $(1)$ yields \begin{align} \sum_{k=1}^{2n}(-1)^k\sqrt{k} &=\sqrt{8}\sum_{k=1}^{n}\sqrt{k}\;-\;\sum_{k=1}^{2n}\sqrt{k}\\ &=\sqrt{8}\left(\frac23n^{3/2}+\frac12n^{1/2}+C+\frac{1}{24}n^{-1/2}\right)\\ &-\left(\sqrt{8}\frac23n^{3/2}+\sqrt2{}\frac12n^{1/2}+C+\frac{1}{\sqrt{2}}\frac{1}{24}n^{-1/2}\right)+O\left(n^{-3/2}\right)\\ &=\frac{1}{\sqrt{2}}n^{1/2}+(\sqrt{8}-1)C+\frac{3}{\sqrt{2}}\frac{1}{24}n^{-1/2}+O\left(n^{-3/2}\right)\tag{4} \end{align} Thus, for even $n$, we get $$\sum_{k=1}^{n}(-1)^k\sqrt{k}=\frac12n^{1/2}+\frac18n^{-1/2}+\frac{7C}{\sqrt{8}+1}+O\left(n^{-3/2}\right)\tag{5}$$ For odd $n$, we get \begin{align} \sum_{k=1}^{n}(-1)^k\sqrt{k} &=\sum_{k=1}^{n-1}(-1)^k\sqrt{k}\;-\;\sqrt{n}\\ &=\color{red}{\frac12(n-1)^{1/2}}+\color{green}{\frac18(n-1)^{-1/2}}-n^{1/2}+\frac{7C}{\sqrt{8}+1}+O\left(n^{-3/2}\right)\\ &=\color{red}{\frac12n^{1/2}-\frac14n^{-1/2}}+\color{green}{\frac18n^{-1/2}}-n^{1/2}+\frac{7C}{\sqrt{8}+1}+O\left(n^{-3/2}\right)\\ &=-\frac12n^{1/2}-\frac18n^{-1/2}+\frac{7C}{\sqrt{8}+1}+O\left(n^{-3/2}\right)\tag{6} \end{align} using the $\color{red}{\text{Binomial}}$ $\color{green}{\text{Theorem}}$ to expand the powers of $(n-1)$ to $O\left(n^{-3/2}\right)$.

Combining $(5)$ and $(6)$, we get $$\sum_{k=1}^{n}(-1)^k\sqrt{k}=(-1)^n\left(\frac12n^{1/2}+\frac18n^{-1/2}\right)-\frac{7}{\sqrt{8}+1}\frac{\zeta(\frac32)}{4\pi}+O\left(n^{-3/2}\right)\tag{7}$$ We've used more powerful machinery, but we've also gotten a more precise answer.

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Thank you for this detailed answer! –  Chon Jun 7 '12 at 19:44
@Chon: I've expanded the answer to use more terms of the Euler-Maclaurin Sum Formula. This reveals the usefulness of that formula. –  robjohn Jun 7 '12 at 20:00