# Two divergent series whose minimum converges?

## Main Question or Discussion Point

I am struggling to find two divergent series, $$\Sigma$$an and $$\Sigma$$bn, such that the series of minimum terms, $$\Sigma$$min{an,bn}, actually converges.

A further stipulation is that both an and bn must be positive, decreasing sequences. (Otherwise the problem is trivial, as one could simply alternate 1/n and 1/n^2 to achieve the desired result.)

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Here's one way. For concision I'm just going to list the reciprocals of the terms:

Code:
{a_n} = 1  4  4 16  25  25  25  25  25  25   25 144 169 196 ...

{b_n} = 1  1  9  9   9  36  49  64  81 100  121 121 121 121 ...
Clearly these are positive and decreasing, and the series of minimum terms is the square harmonic series, which converges. But we let the two series take increasingly long "turns" being constant, long enough that each can be seen to be at least as large as the harmonic series.

For instance, the bottom series needs to stay long enough at 1/121 so that the part after 9,9,9 (which corresponds to 1/3) sums to at least (1/4).

Thanks, boss!

I had attempted a pair that was similar, but too simple and not quite workable. Here's what I initially tried (in reciprocal):

a_n = 1+9+9+ 9+ 25+25+25+25+25+49+49+49+49+49+49+49+81 … (nine times) …
b_n = 4+4+16+16+16+16+36+36+36+36+36+36+64 … (eight times) …

Then a_n is (2k+1)/(2k+1)^2 = 1/(2k+1), b_n is 2k/(2k) ^2 = 1/2k,

but the sequence of minimum terms is [k/2]/k^2 (where [k/2] denotes the floor function), which is approximately 1/2k.

Your sequences seem to be neatly-defined, letting each "go out far enough" to get the next term in the (basic) harmonic series. I would be hard pressed to find a closed-form expression for them, although I'd think it would be easy to define them algorithmically.

arildno