# Proof

It is sufficient to find only one counter-example of two convergent series $$\sum_{n=0}^\infty a_n$$ and $$\sum_{n=0}^\infty b_n$$, the Cauchy product of which does not converge.

We construct such a counter-example. For a fixed $$n$$ set

$a_n:=b_n:=\frac{(-1)^n}{\sqrt{n+1}}\quad\text{and}\quad c_n:=\sum_{k=0}^n a_{n-k}b_k.$

We will first show that the series $$\sum_{n=0}^\infty a_n$$ and $$\sum_{n=0}^\infty b_n$$ are convergent. This follows from the criterion for alternating infinite series, because the sequence $\left(\frac{1}{\sqrt{n+1}}\right)_{n\in\mathbb N}$ is a monotonically decreasing real sequence, which converges to $$0$$.

The Cauchy product has the terms $c_n:=\sum_{k=0}^n a_{n-k}b_k=\sum_{k=0}^n \frac{(-1)^{n-k}}{\sqrt{n-k+1}}\cdot\frac{(-1)^n}{\sqrt{n+1}}=(-1)^n\sum_{k=0}^n\frac 1{\sqrt{(n-k+1)(k+1)}}.$ Now, for all $$k=0,\ldots,n$$ we can estimate $(n-k+1)(k+1)=(n-k)k+(n-k)+(k+1)=(n-k)k+(n+1)\le n^2+(n+1) < n^2+2n+1=(n+1)^2,$ and $\sqrt{(n-k+1)(k+1)} < (n+1).$ and thus, for all $$k=0,\ldots,n$$, we get the estimation $\frac {1}{\sqrt{(n-k+1)(k+1)}}>\frac {1}{(n+1)}.$ With this estimation, we can therefore finally estimate
$|c_n|=\left|\sum_{k=0}^n\frac {1}{\sqrt{(n-k+1)(k+1)}}\right| > \left|\sum_{k=0}^n\frac {1}{(n+1)}\right|=(n+1)\cdot \frac {1}{(n+1)}=1.$ This demonstrates that the series $$\sum_{n=0}^\infty c_n$$ cannot converge, since the absolute value of every term is greater then $$1$$. (If the series converged, then it would violate the necessary condition for convergent series requiring all terms to converge to $$0$$.)

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983