The product of two divergent series is divergent? This is an TRUE/FALSE queston:
The product of two divergent series is divergent.
The correct answer is FALSE.
I know that the product of two convergent series may not be convergent (i.e. $\frac{(-1)^n}{\sqrt{n}}$) according to Cauchy Product. 
My question is why "The product of two divergent series may not be divergent"?? Is there any counter example?
Thanks!
 A: Assuming you mean pointwise product, this is the simplest counterexample I could think of: $$\sum_{n=1}^\infty \left( \frac{1}{n} \right) \!\!\! \left( \frac{1}{n} \right) = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
but $\sum_{n=1}^\infty \frac{1}{n}$ diverges.
A: Take the series $(-1,-2,-2,-2,\dots)$ and the series $(-1,2,-2,2,-2,\dots)$. The Cauchy product of these two series is $(1,0,0,0,\dots)$.
You can think of these as the power series for the  the function $$f(z)=\frac{z+1}{z-1} = 1-2\sum_i z^i$$ and $$g(z)=\frac{z-1}{z+1}=1-2\sum_{i} (-1)^iz^i=f(-z)$$ Neither power series converges at $z=1$, which is the statement that $\sum a_i$ and $\sum b_i$ diverges, but their Cauchy product gives the identity power series...
Essentially, the underlying operation of Cauchy products is the product of the corresponding power series. (This is the "obvious" reason why the Cauchy product of sequences is associative, for example...)
You can define $a_k=k^k$, (with $a_0=1$) and the (Cauchy) multiplicative inverse can be found and the power series likewise will have radius of convergence $0$, and thus $b_i\to\infty$ too.
