Series $\displaystyle \sum\frac{1}{b_n}$ where $\displaystyle b_n=\sum_{k=1}^n(-1)^{k+1}\sqrt{k}.$ Let 
$$b_n=\sum_{k=1}^n(-1)^{k+1}\sqrt{k}.$$
I look for the following series $\displaystyle \sum\frac{1}{b_n}$ is convergent or divergent ? Any suggestion ?
I am assuming $$b_n = \sum_{k=1}^n (-1)^{k+1} \sqrt{k}$$
We have
\begin{align}
b_{2n} &amp; = \left(\sqrt{1} + \sqrt{3} + \sqrt{5} + \cdots +\sqrt{2n-1} \right) - \left(\sqrt2 + \sqrt4 + \sqrt6 + \cdots + \sqrt{2n}\right)\\
&amp; = c_{2n} - 2\sqrt2 c_n
\end{align}
\begin{align}
b_{2n+1} &amp; = \left(\sqrt{1} + \sqrt{3} + \sqrt{5} + \cdots +\sqrt{2n-1} + \sqrt{2n+1}\right) - \left(\sqrt2 + \sqrt4 + \sqrt6 + \cdots + \sqrt{2n}\right)\\
&amp; = c_{2n+1} - 2\sqrt2 c_n
\end{align}
where $c_n = \sqrt1 + \sqrt2 + \cdots + \sqrt{n}$. Now $c_n \sim \dfrac23 {n^{3/2}} + \dfrac12 {n^{1/2}} + \dfrac1{24} \cdot \dfrac1{n^{1/2}}$. Hence, we have
\begin{align}
b_{2n} &amp; \sim \left(\dfrac23 {(2n)^{3/2}} + \dfrac12 {(2n)^{1/2}} + \dfrac1{24\sqrt2} \cdot \dfrac1{n^{1/2}}\right) - 2\sqrt2 \left(\dfrac23 {n^{3/2}} + \dfrac12 {n^{1/2}} + \dfrac1{24} \cdot \dfrac1{n^{1/2}}\right)\\
&amp; = - \dfrac1{\sqrt2} n^{1/2} - \dfrac1{8 \sqrt2} \cdot \dfrac1{n^{1/2}}
\end{align}
\begin{align}
b_{2n+1} &amp; \sim \left(\dfrac23 {(2n+1)^{3/2}} + \dfrac12 {(2n+1)^{1/2}}\right) - 2\sqrt2 \left(\dfrac23 {n^{3/2}} + \dfrac12 {n^{1/2}} \right) - \dfrac1{8 \sqrt2} \cdot \dfrac1{n^{1/2}}\\
&amp; = \left(\dfrac23 (2n)^{3/2} \left(1 + \dfrac1{2n}\right)^{3/2} + \dfrac12 (2n)^{1/2} \left(1+\dfrac1{2n} \right)^{1/2} \right) - 2\sqrt2 \left(\dfrac23 {n^{3/2}} + \dfrac12 {n^{1/2}} \right) - \dfrac1{8 \sqrt2} \cdot \dfrac1{n^{1/2}}\\
&amp; \sim \dfrac{4 \sqrt2}3n^{3/2} + \dfrac{4 \sqrt2}3n^{3/2} \cdot \dfrac3{4n} + \dfrac{4 \sqrt2}3n^{3/2} \cdot \dfrac{\dfrac32 \cdot \dfrac12}2 \cdot \dfrac1{4n^2} + \dfrac{\sqrt2}2 \cdot n^{1/2} + \dfrac{n^{1/2}}{\sqrt2} \cdot \dfrac1{4n} - \dfrac{4 \sqrt2}3 \cdot n^{3/2} - \sqrt2 \cdot n^{1/2} - \dfrac1{8 \sqrt2} \cdot \dfrac1{n^{1/2}}\\
&amp; =  \dfrac1{\sqrt2} n^{1/2} + \dfrac3{8\sqrt2} \cdot \dfrac1{n^{1/2}}
\end{align}
Hence,
\begin{align}
\dfrac1{b_{2n}} + \dfrac1{b_{2n+1}} &amp; \sim - \dfrac{\sqrt2}{n^{1/2} + \dfrac18 \cdot \dfrac1{n^{1/2}}} + \dfrac{\sqrt2}{n^{1/2}+ \dfrac38 \cdot \dfrac1{n^{1/2}}}\\
&amp; = -\sqrt2 \dfrac{\left( \dfrac38 \cdot \dfrac1{n^{1/2}} - \dfrac18 \cdot \dfrac1{n^{1/2}}\right)}{\left( n^{1/2} + \dfrac18 \cdot \dfrac1{n^{1/2}}\right)\left( n^{1/2} + \dfrac38 \cdot \dfrac1{n^{1/2}}\right)}\\
&amp; = \dfrac{-\dfrac1{2\sqrt2}}{n^{3/2} + \dfrac12 \cdot n^{1/2} + \dfrac3{64} \cdot \dfrac1{n^{1/2}}} \sim - \dfrac1{(2n)^{3/2}}
\end{align}
Hence,
$$\sum_{k=1}^{\infty} \dfrac1{b_k} = \dfrac1{b_1} + \sum_{k=1}^{\infty} \left(\dfrac1{b_{2k}} + \dfrac1{b_{2k+1}} \right) \sim 1 - \sum_{n=1}^{\infty} \dfrac1{(2n)^{3/2}} \text{ which converges}$$
The series $b_n$ is not convergent, so $\sum_{k=1}^\infty \frac{1}{b_n}$ is probably convergent. Though you would have to use some sort of test, maybe ratio or integral or something of the sort.

EDIT: I interpreted $b_n$ with $\sqrt k$ and not $\sqrt n$, as I'm not totally sure what you're writing.