mersenneforum.org Miscellany log-series question
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 2022-02-19, 21:32 #1 mart_r     Dec 2008 you know...around... 3×11×23 Posts Miscellany log-series question Just trolling for replies. And maybe this is not really interesting anyway. If I crunched the numbers right, $\sum_{n=2}^x \frac{1}{n \log n} \sim log_2 x + 0.79467864545...$ $\sum_{n=2}^x \frac{1}{n \log n \log_2 n} \sim log_3 x + 2.6957421128...$ $\sum_{n=2}^x \frac{1}{n \log n \log_2 n \log_3 n} \sim log_4 x - 7.5283225139...+0.5684516772...i$ $\sum_{n=2}^x \frac{1}{n \log n \log_2 n \log_3 n \log_4 n} \sim log_5 x - 9.2121545975...+2.0048113305...i$ (Maybe start the third sum with n=3 and the fourth with n=16 to keep the numbers real, in which case the constants would be -7.7099396857... and -10.7320827413... respectively.) Haven't found anything about this series yet. Is it unknown? Should I continue? Also, has the difference between $\sum_{n=2}^x \frac{1}{\log n^s}\quad \mbox{ and } \quad \int_{n=2}^x \frac{1}{\log n^s}$ ever been looked at? Something like the $$\zeta(s)$$ for reciprocal logs?
 2022-02-20, 13:05 #2 Dobri   "Καλός" May 2018 17×19 Posts Note that Log[n^s] = s*Log[n], and Sum[1/Log[n^s],{n,2,x}] = (1/s)*Sum[1/Log[n],{n,2,x}] where Sum[1/Log[n],{n,2,x}] diverges for x -> Infinity. Also, Integrate[1/Log[n^s],{n,2,x}] = (li[x] - li[2])/s where li is the logarithmic integral function, see https://en.wikipedia.org/wiki/Logari...egral_function.
2022-02-20, 18:09   #3
mart_r

Dec 2008
you know...around...

13678 Posts

Quote:
 Originally Posted by Dobri Also, Integrate[1/Log[n^s],{n,2,x}] = (li[x] - li[2])/s where li is the logarithmic integral function, see https://en.wikipedia.org/wiki/Logari...egral_function.
I'm terribly sorry, I meant the difference between

$\sum_{n=2}^x \frac{1}{(\log n)^s} - \int_{2}^x \frac{dn}{(\log n)^s}$
which should constitute an entire function for $$\lim x \to \infty$$ and $$s \in \mathbb C^+$$.

 2022-02-21, 14:58 #4 Dr Sardonicus     Feb 2017 Nowhere 133008 Posts I suggest you look up:Integral test for convergence Euler-Maclaurin summation formula As to the series with general terms 1/n^s, 1/(n*(log(n))^s), 1/(n*log(n)*(log(log(n))^s) etc these are standard examples of using the integral test. I note that log(n) > 0 for n > e^0 = 1; log(log(n)) > 0 for n > e^1 = e; log(log(log(n))) > 0 for n > e^e, etc. The constant associated with the "harmonic series," $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k} - \log(n)$ is called Mascheroni's, or the Euler-Mascheroni constant. It is usually denoted by $\gamma$.
2022-02-22, 20:56   #5
mart_r

Dec 2008
you know...around...

3·11·23 Posts

Quote:
 Originally Posted by Dr Sardonicus I suggest you look up:Integral test for convergence Euler-Maclaurin summation formula
Aha, that's exactly the keywords I was looking for.
I've seen this before (though haven't taken a closer look at it before), but didn't draw the line to the problem at hand.

Thanks so much for the replies! I was already in serious worry that many forum members have me on their ignore list for whatever reason.

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