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?



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.

  • $\begingroup$ Thanks a lot! But I am still confused. According to Cauchy Product which says 'at least one series is absolute convergent then the product is convergent'. So why the product of two divergent series may convergent? I mean does it disobey the theorem? $\endgroup$ – 52145208 Jun 15 '14 at 1:36
  • $\begingroup$ @52145208: That article only states a 1-directional implication (and you've left off half the statement). You can't say that if the product converges, then both factors have to converge; that would be an application of the theorem's converse, but the theorem's converse isn't true. $\endgroup$ – user2357112 Jun 15 '14 at 1:59
  • $\begingroup$ @user2357112 Okay. But is there any "elementary" example of this question? I cannot quite follow this answer. $\endgroup$ – 52145208 Jun 15 '14 at 2:13
  • $\begingroup$ Not clear what you mean by "I cannot quite follow this answer." You can try it with $(a_i)=(1,2,2,2,2,2,\dots)$ and $(b_i)=(1,-2,2,-2,2\dots)$ and prove directly that the Cauchy product is $(1,0,0,0,\dots)$ $\endgroup$ – Thomas Andrews Jun 15 '14 at 2:49
  • 1
    $\begingroup$ That's not the Cauchy production of the series. $\endgroup$ – Thomas Andrews Jun 15 '14 at 5:06

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.

  • 3
    $\begingroup$ For a simpler counterexample, take the harmonic series on even / odd integers separately, and $0$ everywhere else. The pointwise product is zero. $\endgroup$ – user61527 Jun 14 '14 at 23:32
  • $\begingroup$ @JoeyBF I did not mean this example. This is not Cauchy Product. Actually this is divergent if calculating by Cauchy Product. $\endgroup$ – 52145208 Jun 14 '14 at 23:34
  • $\begingroup$ @user61527 Can you give me more details? I cannot understand it. Thanks! $\endgroup$ – 52145208 Jun 14 '14 at 23:35
  • $\begingroup$ $a_n=b_n=\frac{1}{n},\ c_n = \sum_{k=0}^n \frac{1}{\sqrt{k+1}} \cdot \frac{1}{\sqrt{n-k+1}} = \sum_{k=0}^n \frac{1}{\sqrt{(k+1)(n-k+1)}},\ \text{Since}\ k+1\le n+1,\ n-k+1\le n+1, \Rightarrow \sqrt{(k+1)(n-k+1)}\le n+1\Rightarrow c_n\ge\sum_{k=0}^{n}\frac{1}{n+1}=1$, so it is divergent. $\endgroup$ – 52145208 Jun 14 '14 at 23:38
  • 1
    $\begingroup$ @ThomasAndrews Sorry? I think you mean to ping 52145208. $\endgroup$ – user61527 Jun 15 '14 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.