Addition and analysing series I need to analyse the convergence of the following and try and prove it is alternating. It says let us denote the sum by S:

$$\sum_{n=1}^\infty c_n = \sum_{n=1}^\infty (b_{n+1} - b_n)  \\\text{where }b_n=a_n +a_{n+1} \text{ and }a_n=\sum_{k=1}^n(-1)^k\sqrt{k} $$

First of all am I correct in saying that:
$$ b_{n+1}-b_n =\\ \Bigr(\sum_{k=1}^{n+1}(-1)^k\sqrt{k} + \sum_{k=1}^{n+2}(-1)^k\sqrt{k}\Bigl) - \Bigr(\sum_{k=1}^{n}(-1)^k\sqrt{k} + \sum_{k=1}^{n+1}(-1)^k\sqrt{k}\Bigl) \text{ =} \\ \Bigr(\sum_{k=1}^{n+2}(-1)^k\sqrt{k}\Bigl) - \Bigr(\sum_{k=1}^{n}(-1)^k\sqrt{k}\Bigl)$$
If so, where do I take it from here? Sorry if I have not phrased this very well.
 A: Your intuition so far is correct. The next thing to notice is that $$\sum_{k=1}^{n+2}m_n = m_{n+2} + m_{n+1} + \sum_{k=1}^{n}m_n$$ and therefore $$\sum_{k=1}^{n+2}m_n - \sum_{k=1}^{n}m_n = m_{n+2} + m_{n+1}$$ This tells us that $$c_n = b_{n+1}-b_n = (-1)^{n+2}\sqrt{n+2} + (-1)^{n+1}\sqrt{n+1} = (-1)^n(\sqrt{n+2}-\sqrt{n+1})$$
Thus we are analyzing the convergence of $$\sum_{n=1}^\infty(-1)^n(\sqrt{n+2}-\sqrt{n+1})$$ Now, we can use the alternating series test to prove that this series is convergent by proving that $\lim_{n\to \infty} |{c_n}| = \lim_{n\to \infty} (\sqrt{n+2}-\sqrt{n+1})$ converges monotonically to 0. One way to do so is to rewrite the limit as $$\lim_{n\to \infty} (\sqrt{n+2}-\sqrt{n+1}) = \lim_{n\to \infty} \frac{1}{\sqrt{n+2}+\sqrt{n+1}}$$ Clearly, this limit approaches 0 monotonically, and therefore our series is convergent by the alternating series test.
A: Yes your claim is correct.
From here:
$$b_{n+1}-b_n =\Big(\sum_{k=1}^{n+2}(-1)^k\sqrt k \Big)-\Big(\sum_{k=1}^{n}(-1)^k\sqrt k \Big)=
 (-1)^{n+2}\sqrt{n+2}+(-1)^{n+1}\sqrt{n+1}= (-1)^{n+2}\big(\sqrt{n+2}-\sqrt{n+1}\big)$$
Now we see that $S$ is alternating. Therefore we must check that $\lim_{n\to \infty}c_n=0$: 
$$\lim_{n\to \infty}(-1)^{n}(\sqrt{n+1}-\sqrt{n})=\lim_{n\to \infty}(-1)^{n}(\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n }=\lim_{n\to \infty}\frac{(-1)^n}{\sqrt{n+1}+\sqrt n }=0$$
and that $\mid c_{n+1} \mid<\mid c_n \mid$:
$$\mid c_n\mid -\mid c_{n+1}\mid=\sqrt{n+1}-\sqrt n- \sqrt{n+2}+\sqrt {n+1}=$$$$\frac{1}{\sqrt{n+1}+\sqrt n}-\frac{1}{\sqrt{n+2}+\sqrt {n+1}}=\frac{\sqrt{n+2}+\sqrt {n+1} -\sqrt{n+1}-\sqrt n}{(\sqrt{n+1}+\sqrt n)(\sqrt{n+2}+\sqrt {n+1})}>0$$
Therefore series is convergent.
