# Multiplying convergent and divergent sequences

When testing for null sequences I've had to say whether they were convergent or divergent, but say you've got a convergent sequence (a) and divergent sequence (b) and you multiplied them (so {ab}) would it make a divergent sequence or would it just cancel?

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In general, you can't say anything about the convergence properties of a sequence $(a_nb_n)$ if one of the sequences $(a_n)$ or $(b_n)$ diverges, even if one of them converges to 0.

Some examples:

• Take $a_n=(-1)^n$ and $b_n=(-1)^{n+1}$. Both of $(a_n)$ and $(b_n)$ diverge, but the sequence $(a_nb_n)=(-1)$ converges.
• Take $a_n=n$ and $b_n={1\over n^2}$. Then $(a_n)$ diverges, $(b_n)$ converges to $0$, but the sequence $(a_n b_n)=({1\over n})$ converges.
• Take $a_n=n^2$ and $b_n={1\over n}$. Then $(a_n)$ diverges, $(b_n)$ converges to $0$, but the sequence $(a_n b_n)=({ n})$ diverges.

In the third example above, though one sequence is converging to zero, the other sequence is tending to infinity fast enough that the "product sequence" tends to infinity. In the second example, though one sequence is tending to infinity, the other sequence is heading to zero quickly enough that the product sequence tends to zero.

Another point to be made: Even if you know that $(a_n)$ diverges, $(b_n)$ converges to $0$, and that $(a_nb_n)$ converges, you can not conclude that $(a_nb_n)$ converges to $0$. For example, take $a_n=n$ and $b_n={3\over n}$.

Here is one positive result though:

• If $(a_n)$ diverges but is bounded and if $(b_n)$ converges to $0$, then $(a_nb_n)$ converges to $0$.
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What if $\lim_{n\to\infty} x_n = \infty$ and $\lim_{n\to\infty} y_n = y > 0$. It appears that we can say $\lim_{n\to\infty} x_n y_n = \infty$. The case for when $y < 0$ is similar, but we end up with $\lim_{n\to\infty} x_n y_n = -\infty$. The proof seems straightforward enough, but I haven't seen a statement like this in a book. Thoughts? – animal_magic May 1 at 18:44