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1+1/2+1/4+1/6+... is a counter example. Proof: 1/4+1/6+1/8 > 1/2

In general, the larger n is, in 1+1/2+1/(2+n)+1/(2+2n)+..., the slower the divergence.

EDIT: Even more general, the faster the arithmetic series in the denominator diverges, the slower the series diverges.

This is an interesting question. Congrats.

In general, the larger n is, in 1+1/2+1/(2+n)+1/(2+2n)+..., the slower the divergence.

EDIT: Even more general, the faster the arithmetic series in the denominator diverges, the slower the series diverges.

This is an interesting question. Congrats.

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AlephZero

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2 terms (1/3 + 1/4) sum to > 1/2

the next 4 terms sum to > 1/2

the next 8 terms sum to > 1/2

and so on for blocks of 2^k terms.

You could construct a slower divergiing series, for example one where

2 terms sum to > 1/3

4 terms sum to > 1/4

8 terms sum to > 1/5

16 terms sum to > 1/6

and so on.

And then construct an infinite set of series, each diverging slower than the previous one, by repeating this process.

Or to put it another way, the first N terms of the harmonic series sums to approximately log(N), but you could construct slower diverging series where the first N terms sum to approximately log(log(N)), log(log(log(N))), and so on.

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The reason I was actually asking was because I was looking at the infinite series of 1/3

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lurflurf

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1/2+1/3+1/5+1/7+1/11+1/13+1/17+1/19+...+1/n~log log n

where each denominator is prime

compare to the harmonic series

1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+...+1/n~log n

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