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I faced a series question it goes something like give an example of 2 divergent series such that when the 2 series are multiplied to each other, the new series becomes convergent, although it looks absurdly simple still am at a loss.

Arnav Das
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3 Answers3

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It depends on what you mean by multiply.. If you're looking for series $a_n$, $b_n$ such that

$\sum a_n$ diverges, $\sum b_n$ diverges, but $\sum a_nb_n$ converges, then you can just take $a_n = \frac 1n = b_n$.

Another way to intend multiplication is to use Cauchy's product, but that requires that the two series to converge and at least one of them do so absolutely, so I guess this is not what you're talking about :-)

Edit:

What I said about the cauchy product is not correct: If those conditions are met, then the product converges, but it's not a necessary condition. Refer to Daniel Fisher comment below ;-)

Ant
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  • Really am not actually able to get you.....according to my book divergence occurs when n tends to infinity then the series should too tend to infinity but as you advised to take both the series as 1\n it doesn't happen here right since as n tends to infinity 1\n tends to 0 right.....please correct me if am somewhere wrong – Arnav Das Jul 05 '15 at 08:17
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    @ArnavDas Do not confuse a series with it's general term. The general term of $\frac 1n$ indeed goes to $0$, but the sum $1 + 1/2 + 1/3 + \dots$ does not! A necessary condition for the series to converge is that it's term goes to $0$; that is to say, if the general term tends to infinity or to some other value different than $0$, then the series diverges. The converse is not true; even is the $n$-th term tends to $0$, the series can still diverge (and the classic example is $\sum \frac 1n$). – Ant Jul 05 '15 at 08:20
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    1/n tends to 0, but Σ1/n tends to $\infty$. – Bernard Jul 05 '15 at 08:20
  • Oh I am really sorry I just missed out the summation symbol – Arnav Das Jul 05 '15 at 08:26
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    The Cauchy product of two divergent series can be convergent. Consider the Taylor series of $f$ and $g$, where $f$ has a simple pole at $1$ and a zero at $-1$, and $g$ has a simple pole at $-1$ and a zero at $1$, and neither has any further poles in the closed unit disk. Then the series of $f$ cannot converge for $z = 1$, and the series of $g$ need not converge for $z = 1$, but the Cauchy product is the Taylor series of $f\cdot g$, and that has a radius of convergence $> 1$, so converges for $z = 1$. – Daniel Fischer Jul 05 '15 at 10:22
  • @DanielFischer Oh thank you for the correction! I had overlooked that :D – Ant Jul 05 '15 at 17:08
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A pair of divergent sequences $$a_n = \begin{cases}0 & \text{for } n \text{ odd} \\ 1 & \text{for } n \text{ even}\end{cases} \quad \text{ and } \quad b_n = 1-a_n$$ makes as a 'product' a convergent sequence $$a_n\cdot b_n = 0$$ and similary their series: while $\sum\limits_n{a_n} = \sum\limits_n{b_n} = \infty$ the 'product' is $$\sum\limits_n{a_n\cdot b_n} = 0$$

CiaPan
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$a_n = (-1)^n = b_n$. Both are divergent but product is {1,1,1....} clearly converges to 1.

Agniswar Chakraborty
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