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Old 2022-02-19, 21:32   #1
mart_r
 
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Default 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?
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Old 2022-02-20, 13:05   #2
Dobri
 
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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.
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Old 2022-02-20, 18:09   #3
mart_r
 
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Quote:
Originally Posted by Dobri View Post
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 \(s \in \mathbb C^+\).
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Old 2022-02-21, 14:58   #4
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I suggest you look up:
  1. Integral test for convergence
  2. 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.
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Old 2022-02-22, 20:56   #5
mart_r
 
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Quote:
Originally Posted by Dr Sardonicus View Post
I suggest you look up:
  1. Integral test for convergence
  2. 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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