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Harmonic series, slowest diverging series? 
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#1
Jan1111, 05:20 AM

P: 398

The harmonic series is divergent, and in general, I know that just because one series is larger than another divergent series, doesn't mean the series is convergent. However, the harmonic series is very, very slow to diverge. Is the harmonic series the slowest diverging series? That is, is it the case that any series larger than the harmonic series must necessarily converge and is there any easy proof to show this?



#2
Jan1111, 06:23 AM

P: 737

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. 


#3
Jan1111, 02:51 PM

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The harmonic series is "exponentially slow" to diverge, in the sense that
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. 


#4
Jan1111, 07:19 PM

P: 398

Harmonic series, slowest diverging series?
Thanks, yeah, that makes sense.
The reason I was actually asking was because I was looking at the infinite series of 1/3^{ln(n)} and recognized it to be smaller than the harmonic series, which would 1/e^{ln(n)}. I was supposed to determine whether that first series converged or diverged and I couldn't figure out how to determine it, but I wondered, since it was smaller than the harmonic series, if it was necessarily convergent. I know now that it doesn't, but if you had an infinite series of the form 1/k^{ln(n)}, for what k values is the sum convergent and why? 


#5
Feb2111, 03:26 AM

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The idea of a least diverget series is doomed. There will always be ways to slow such a series much more. Like AlephZero mentioned there is the famous prime harmonic series
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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