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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 ?

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Is it $\sqrt n$ or $\sqrt k$? –  Julián Aguirre Mar 1 '13 at 17:32
    
Sorry, it's $\sqrt {k}$. –  Sami Ben Romdhane Mar 1 '13 at 18:07
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I am assuming $$b_n = \sum_{k=1}^n (-1)^{k+1} \sqrt{k}$$ We have \begin{align} b_{2n} & = \left(\sqrt{1} + \sqrt{3} + \sqrt{5} + \cdots +\sqrt{2n-1} \right) - \left(\sqrt2 + \sqrt4 + \sqrt6 + \cdots + \sqrt{2n}\right)\\ & = c_{2n} - 2\sqrt2 c_n \end{align} \begin{align} b_{2n+1} & = \left(\sqrt{1} + \sqrt{3} + \sqrt{5} + \cdots +\sqrt{2n-1} + \sqrt{2n+1}\right) - \left(\sqrt2 + \sqrt4 + \sqrt6 + \cdots + \sqrt{2n}\right)\\ & = 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} & \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)\\ & = - \dfrac1{\sqrt2} n^{1/2} - \dfrac1{8 \sqrt2} \cdot \dfrac1{n^{1/2}} \end{align} \begin{align} b_{2n+1} & \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}}\\ & = \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}}\\ & \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}}\\ & = \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}} & \sim - \dfrac{\sqrt2}{n^{1/2} + \dfrac18 \cdot \dfrac1{n^{1/2}}} + \dfrac{\sqrt2}{n^{1/2}+ \dfrac38 \cdot \dfrac1{n^{1/2}}}\\ & = -\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)}\\ & = \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, $$\sum_{k=1}^{\infty} \dfrac1{b_k} = \dfrac1{b_1} + \sum_{k=1}^{\infty} \left(\dfrac1{b_{2k}} + \dfrac1{b_{2k+1}} \right) \sim 1 - \sum_{n=1}^{\infty} \dfrac1{(2n)^{3/2}} \text{ which converges}$$

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Sorry, but we can not use comparaison test to a series with alternating terms, is not it? –  Sami Ben Romdhane Mar 1 '13 at 18:23
    
@sbr Thanks. Have now updated it. –  user17762 Mar 1 '13 at 19:05
    
Can you please explain how you calculate the expension of $c_n$ ? –  user10676 Mar 1 '13 at 20:41
    
@Marvis Good job (+1). Thanks. –  Sami Ben Romdhane Mar 1 '13 at 20:54
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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.

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the integral test won't work as the series is not absolut convergent, and it isn't monotone anyway ... –  Dominic Michaelis Mar 1 '13 at 18:12
    
@dominic you're right. I don't think you can say much about this series without writing something down, I think. –  noobProgrammer Mar 1 '13 at 20:12
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