Thanks Peter! It's really hard to think an counter example for this statement. I searched lots of articles and textbooks of calculus and analysis yet there is no result.

Depends a little upon what you mean by "divergent". For example, one answer might be the Taylor series of the two functions sqrt(1-x) and 1/sqrt(1-x) which are divergent for |x|>1. However, their product is just 1 for all x and hence is convergent for all x.

Alternatively you might want "divergent" to mean a divergent power series like the Stieltjes series, \( f(x):=\sum_{n=1} n!(-x)^n \), which diverges for all |x|>0. Multiply by \( g(x):=1/f(x)=1+x-x^2+3x^3-13x^4+71x^5-461x^6+3447x^7 +\cdots\) which I believe is everywhere divergent, and the result is again 1 with an everywhere convergent series.

If only one series is assumed to be divergent that is easy. Take (1-z) and 1/(1-z) in z=1.

Thanks to all!

This question is from a Calculus-1 course in China. They do not teach Taylor series.

In my opinion, there are two types of meaning of this question.

One is "pairwise product", for instance, a_n=b_n=1/n which is a harmonic series (divergent) and their pairwise product is 1 / n^2 which is convergent.

Another type of this question is Cauchy product. I cannot find a suitable counter example in the second type (Cauchy Product).

Thanks again!

The pairwise product is of no interest to higher level mathematics.

For the Cauchy product without Taylor series just use whatever you get from the Taylor series for sqrt(1-x) and 1/sqrt(1-x) but 'evaluated' at x=2 so they are both divergent. The two series have easily known analytic expressions for each term. The Cauchy product of the series is 1+0+0+0+... which is convergent.

DEar Yin Zhao, I think I have a counter example: consider power series

1/f=(1-z)(1+z) and 1/g=(1-z)(1+z^2). Now each of these power series is divergent in z=1 but their ratio is convergent. Note that the Cauchy product of the Taylor series of f and g sums to the product f(z)g(z). Of course some familiarity with power series might help...

@Peter: I dont understand your counterexample as the series of general term 1 is certainly divergent.

@Peter I think you are mixing convergence of series and convergence of sequences.

all one is convergent as a sequence but not as a series

I understand that by pairwise product of \sum u_n by \sum v_n he means the series

of general term u_nv_n as opposed to Cauchy product \sum_k u_k v_{n-k}. So it makes sense to use power series when looking for a counter example. And convergence of \sum_n u_n x^n at x=1 is equivalent to the convergence of the series

\sum_n u_n by definition.

@Peter: thanks your example is simpler and more convincing than mine. Its a language problem: in french series NEVER mean sequence. Always the partial sum sequence...