Why is the Cauchy product of the series $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ diverging?

Question

I'm new to series and I'm reading old questions here to see some examples of tests for divergence and convergence. In the following, I have a question about the conclusion: https://math.stackexchange.com/questions/279890/prove-the-divergence-of-cauchy-product-of-convergent-series-a-n-b-n-dfra#:~:text=To%20show%20that%20the%20Cauchy,2x%2Cy%E2%88%88R.&text=and%20n%E2%88%91k%3D0,0%2C%20and%20the%20series%20diverges

I have the Cauchy product of the convergent series $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ which gives me:

$\sum_{n=0}^{\infty}\sum_{k=0}^{n}\dfrac{(-1)^{n-k}}{\sqrt{n-k+1}}\cdot \dfrac{(-1)^{k}}{\sqrt{k+1}} = \dfrac{(-1)^n}{\sqrt{n-k+1} \cdot \sqrt{k+1}}$.

Additionally, the following is true:

$\sum_{k=0}^n\frac{1}{\sqrt{n-k+1}\,\sqrt{k+1}}\ge\frac{2(n+1)}{n+2}$

I understand how to get here. My question is this: Why does this inequality mean that the Cauchy product diverges? What happened to the $(-1)^n$?

Answer 1

Well, yes, there is a sign missing. In fact$$\sum_{k=0}^na_kb_{n-k}=(-1)^nc_n,$$where$$c_n\geqslant\frac{2n+2}{n+2}.$$But then you don't have $\lim_{n\to\infty}(-1)^nc_n=0$, and therefore the series $\sum_{n=0}^\infty(-1)^nc_n$ diverges.