Proving a series is not convergent I have an exercise where I have to show that a series is not convergent. I have tried some convergence tests but I'm unfamiliar with working a product in a series.
Let $a_n = b_n = \frac{(-1)^n}{\sqrt{n+1}}$
Now let $c_n = \sum_{k=0}^n a_{n-k}b_k$
Prove that $\sum_{k=0}^\infty c_n$ is not convergent.
As before mentioned. I haven't really worked with series like these. It stated that the Cauchy product formula is wrong here. So I've tried to just calculate the limit using the ratio test, but that did not seem to work.
Is it correct to just write it like this?
$\sum_{k=0}^\infty c_n$ $=$ $\sum_{k=0}^\infty \frac{(-1)^{n-k+k}}{\sqrt{n-k+1}\sqrt{k+1}}$ $=$ $\sum_{k=0}^\infty \frac{(-1)^{n}}{\sqrt{n-k+1}\sqrt{k+1}}$
I'm not sure if I can rewrite $\sqrt{n-k+1}\sqrt{k+1}$ $=$ $\sqrt{(k+1)(-k+n+1)}$ But I'm equally unsure if this is even helps.
Now my question is: Is my work this far even correct or am I making crucial mistakes even just writing these down and if it is, how can I proceed to prove that this series does in fact not converge? (Since the ratio tests was inconclusive)
 A: We'll make use of the following theorems:

*

*$\lim\limits_{n\to\infty}a_n=L$ if and only if $\lim\limits_{n\to\infty}a_{2n}=L$ and $\lim\limits_{n\to\infty}a_{2n+1}=L$.

*If $\lim\limits_{n\to\infty}a_n\neq 0$ or does not exist, then $\sum_{n=0}^{\infty}a_n$ is divergent

Combining both theorems, it follows that if either $\lim_{n\to\infty}c_{2n}$ or $\lim_{n\to\infty}c_{2n+1}$ doesn't exist, then $\lim\limits_{n\to\infty}c_{n}$ doesn't exist either and hence $\sum_{n=0}^{\infty}c_n$ is divergent. Therefore, we may assume that $c_{2n}$ and $c_{2n+1}$ are both convergent.
We proceed by first analyzing the behavior of $c_{2n}$.
\begin{align*}
c_{2n} &= \sum_{k=0}^{2n}\frac{(-1)^{2n}}{\sqrt{2n-k+1}\sqrt{k+1}}\\
&= \sum_{k=0}^{2n}\frac{1}{\sqrt{2n+1-k}\sqrt{k+1}}
\end{align*}
Note that for $0\leq k\leq 2n$, we have that $k+1\geq 1$ and $1\leq 2n+1-k\leq 2n+1$, so each of the terms in the sum $\sum_{k=0}^{2n}\frac{1}{\sqrt{2n+1-k}\sqrt{k+1}}$ is positive. It follows that $c_{2n}$ is strictly positive and strictly increasing, so the limit of $c_{2n}$ must be greater than zero.
Now we analyze the behavior of $c_{2n+1}$.
\begin{align*}
c_{2n+1} &= \sum_{k=0}^{2n+1}\frac{(-1)^{2n+1}}{\sqrt{2n+1-k+1}\sqrt{k+1}}\\
&= \sum_{k=0}^{2n+1}\frac{-1}{\sqrt{2n+2-k}\sqrt{k+1}}
\end{align*}
For $0\leq k\leq 2n+1$, we have that $k+1\geq 1$ and $1\leq 2n+2-k\leq 2n+2$, so each of the terms in the sum $\sum_{k=0}^{2n+1}\frac{-1}{\sqrt{2n+2-k}\sqrt{k+1}}$ is strictly negative (the $-1$ present in each of the terms has this effect). It follows that $c_{2n+1}$ is strictly negative and strictly decreasing, so the limit of $c_{2n+1}$ must be less than zero.
We have shown that $c_{2n}$ has a positive limit and $c_{2n+1}$ has a negative one. Since no real number can be both positive and negative, it follows that $\lim_{n\to\infty}c_n$ cannot exist, and thus $\sum_{n=0}^{\infty}c_n$ is divergent.
