# prove the divergence of cauchy product of convergent series $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$

i am given these series which converge. $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i solved this with quotient test and came to $-1$, which is obviously wrong. because it must be $0<\theta<1$ so that the series converges. my steps:

$\dfrac{(-1)^{n+1}}{\sqrt{n+2}}\cdot \dfrac{\sqrt{n+1}}{(-1)^{n}} = - \dfrac{\sqrt{n+1}}{\sqrt{n+2}} = - \dfrac{\sqrt{n+1}}{\sqrt{n+2}}\cdot \dfrac{\sqrt{n+2}}{\sqrt{n+2}} = - \dfrac{(n+1)\cdot (n+2)}{(n+2)\cdot (n+2)} = - \dfrac{n^2+3n+2}{n^2+4n+4} = -1$

did i do something wrong somewhere?

and i tried to know whether the cauchy produkt diverges as task says:

$\sum_{k=0}^{n}\dfrac{(-1)^{n-k}}{\sqrt{n-k+1}}\cdot \dfrac{(-1)^{k}}{\sqrt{k+1}} = \dfrac{(-1)^n}{nk+n-k^2+1} = ..help.. = diverging$

i am stuck here how to show that the produkt diverges, thanks for any help!

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You should observe that each of the $n+1$ summands is (up to sign) $\ge \frac1{\sqrt{n+1}}\frac1{\sqrt{n+1}}=\frac1{n+1}$. – Hagen von Eitzen Jan 16 '13 at 7:46
yeah, i found my mistake, i didnot take the absolute value, then it will be $1$ and not -1 – doniyor Jan 16 '13 at 7:51

$\sum_{n=0}^\infty\dfrac{(-1)^n}{\sqrt{n+1}}$ is convergent by Leibniz's test, but it is not absolutely convergente (i.e. it is conditionally convergent.)

To show that the Cauchy product does not converge use the inequality $$x\,y\le\frac{x^2+y^2}{2}\quad x,y\in\mathbb{R}.$$ Then $$\sqrt{n-k+1}\,\sqrt{k+1}\le\frac{n+2}{2}$$ and $$\sum_{k=0}^n\frac{1}{\sqrt{n-k+1}\,\sqrt{k+1}}\ge\frac{2(n+1)}{n+2}.$$ This shows that the the terms of the Cauchy product do not converge to $0$, and the series diverges.

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Here is another solution: Observe that

\begin{align*} \sum_{k=0}^{n}\frac{(-1)^{k}}{\sqrt{k+1}} \frac{(-1)^{n-k}}{\sqrt{n-k+1}} &= (-1)^{n} \sum_{k=0}^{n}\frac{1}{\sqrt{(k+1)(n-k+1)}} \\ &= (-1)^{n} \sum_{k=0}^{n}\frac{1}{n} \frac{1}{\sqrt{\left(\frac{k+1}{n}\right)\left(1-\frac{k-1}{n}\right)}}. \end{align*}

But since

$$\sum_{k=0}^{n}\frac{1}{n} \frac{1}{\sqrt{\left(\frac{k+1}{n}\right)\left(1-\frac{k-1}{n}\right)}} \xrightarrow[]{n\to\infty} \int_{0}^{1} \frac{dx}{\sqrt{x(1-x)}} = \pi,$$

the general term of the Cauchy product diverges. Therefore the series itself diverges. (Here, the specific value of the integral is irrelevant. It suffices to observe that the integral is positive.)

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