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I was looking at the counterexample $a_n = b_n = (-1)^n/\sqrt{n}$ where $\sum a_n$ and $\sum b_n$ converge but the Cauchy product $\sum_{k=0}^\infty c_k = \sum_{k=0}^\infty \sum_{j=0}^{k}a_j b_{k-j}$ diverges.

I became curious if it is possible for two series to diverge but have a convergent Cauchy product. After some searching I always found same example using power series that diverge at the radius of convergence, giving

$$a_n: (-1,-2,-2,-2, \ldots)\\b_n: (-1,2,-2,2,-2, \ldots)\\c_n: (1,0,0,0, \ldots)$$

Can anyone show me a less trivial example where the Cauchy product of diverging series is converging but with $c_n \neq 0$ for all $n$ or a subsequence $c_{n_i} \neq 0$. Thanks.

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2 Answers 2

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Such an example can be constructed from a divergent geometric series.

With $\alpha > \beta > 1$ take

$$a_n = \begin{cases}\,\,\,\,\,\,\,\,\,1 \, , \,\,\,\,\,\,\,\,\,\ n= 0 \\ -\left(\frac{\alpha}{\beta}\right)^n \, , \,\,\, n \geqslant1 \end{cases}$$

and

$$b_n = \begin{cases}\,\,\,\,\,1 \, , \,\,\,\,\,\,\,\,\,\ n= 0 \\ \left(\frac{\alpha}{\beta}\right)^{n-1}\left(\beta^n + \beta^{-(n+1)} \right) \, , \,\,\, n \geqslant1 \end{cases}$$

Since $\alpha/\beta > 1$, the series $\sum a_n$ is a divergent geometric series and $\sum b_n$ diverges by the comparison test.

The general term of the Cauchy product is

$$c_n = a_0b_n + \sum_{k=1}^{n-1} a_{n-k} b_{k} + a_nb_0 \\ = \left(\frac{\alpha}{\beta}\right)^{n-1}\left(\beta^n + \beta^{-(n+1)} \right)- \sum_{k=1}^{n-1}\left(\frac{\alpha}{\beta}\right)^{n-k} \left(\frac{\alpha}{\beta}\right)^{k-1}\left(\beta^k + \beta^{-(k+1)} \right) - \left(\frac{\alpha}{\beta}\right)^n \\ = \left(\frac{\alpha}{\beta}\right)^{n-1} \left(\beta^n + \beta^{-(n+1)} -\sum_{k=1}^{n-1}\left(\beta^k + \beta^{-(k+1)} \right) - \frac{\alpha}{\beta} \right)$$

The finite geometric sums appearing on the RHS are

$$\sum_{k=1}^{n-1}\beta^k = \frac{\beta - \beta^n}{1 - \beta} = \frac{\beta^n - \beta}{\beta -1 } , \\ \sum_{k=1}^{n-1} \beta^{-(k+1)} = \frac{\beta^{-2} - \beta^{-(n+1)}}{1 - \beta^{-1}}= \frac{\beta^{-1} - \beta^{-n}}{\beta - 1}$$

Hence,

$$c_n = \left(\frac{\alpha}{\beta}\right)^{n-1} \left(\beta^n + \beta^{-(n+1)} -\frac{\beta^n - \beta + \beta^{-1} - \beta^{-n}}{\beta -1 } - \frac{\alpha}{\beta} \right)$$

Choosing $\beta = 2$ for convenience we get

$$c_n = \left(\frac{\alpha}{2}\right)^{n-1} \left(2^n + 2^{-(n+1)} -2^n + 2 - 2^{-1} + 2^{-n} - \frac{\alpha}{2} \right) \\ = \left(\frac{\alpha}{2}\right)^{n-1}\left(\frac{3}{2^{n+1}} + \frac{3 - \alpha}{2} \right) \\ = \frac{\alpha^{n-1}}{2^{2n}}\left(3 + (3 - \alpha)2^n \right)$$

Since we require $\alpha > \beta = 2$ we can choose $\alpha = 3$ to get

$$c_n = \left(\frac{3}{4}\right)^n,$$

and the Cauchy product $\sum c_n$ is a convergent geometric series with non-zero terms.

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  • $\begingroup$ Thanks. I was surprised nobody answered this. What was the motivation to find this example? $\endgroup$
    – SAS
    May 30, 2017 at 23:20
  • $\begingroup$ @SAS: Working with a geometric series is a help since closed form manipulations are feasible. I started with a divergent series and used the factor $\beta^n + \beta^{-(n+1)}$ hoping the term $\beta^{-(n+1)}$ would factor out of the intermediate sum and everything else would cancel. $\endgroup$
    – RRL
    May 31, 2017 at 3:06
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We can also do this by tweaking your example a bit: Let $\sum x_n$ be any convergent series. Define

$$(a_n) = (-1,-2,-2,-2, \ldots) \\(b_n): (-1+x_1,2+x_2,-2+x_3,2+x_4, \ldots).$$

Then both $\sum a_n, \sum b_n$ diverge. The Cauchy product sequence is then

$$\tag 1(c_n) = (1-x_1, -2x_1 - x_2, -2x_1 - 2x_2-x_3, -2x_1 - 2x_2 - 2x_3 - x_4, \dots) = (1-x_1, -2S_1 - x_2, -2S_2-x_3, -2S_3 - x_4, \dots) .$$

Here $S_n = x_1 + x_2 + \cdots + x_n.$

Let's now specify $(x_n)$ as $(1/2, -1/2, 1/4, -1/4,1/8,-1/8, \dots).$ We then have $S_n = 1/2^{(n+1)/2}$ for $n$ odd, $S_n = 0$ for $n$ even. From this and the formula in $(1),$ you can verify that $\sum |c_n|<\infty$ and $c_n\ne 0$ for every $n.$

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