 mersenneforum.org Miscellany log-series question
 Register FAQ Search Today's Posts Mark Forums Read 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)/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)/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," is called Mascheroni's, or the Euler-Mascheroni constant. It is usually denoted by .   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.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post a1call Lounge 1691 2022-05-20 15:04 siegert81 GPU Computing 47 2011-10-14 00:49 grandpascorpion Math 23 2005-01-24 20:11 Orgasmic Troll Math 1 2004-09-13 19:01 Rosenfeld Puzzles 2 2003-07-01 17:41

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