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I am interested in getting more precise partial sums for $$\sum_{x=1}^n \frac{1} {x \log(x)}$$ I understand that this series will diverge as $n \to \infty$. I've only been able to get a rough estimate for a lower bound based on PNT. Using PNT I can deduce that $p_n$ ~ $n \log(n)$. Using Rosser's Theorem I know that $p_n > n\log(n)$. I've looked into it and found: $\sum_{x=1}^n \frac{1} {x \log(x)}$ ~ $\sum_{x=1}^n \frac{1} {p_x} \approx H_n$ ~ $\log(n)$. With this I know that $\sum_{x=1}^n \frac{1} {x \log(x)} > \log(n)$ because it is the sum of their reciprocals and the nth harmonic $H_n - \log(n) = \gamma$ as $n \to \infty$. I did find an old forum post on a similar question about specifically primes: $\sum_{x=1}^n \frac{1} {p_x \log(p_x)}$ but it proved to be inconclusive. I am interested if there is an asymptotic function with some error that better approximates partial sums. I'm looking to graph sums. I know Taylor Series could help but I'm looking for something more succient if possible.

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  • $\begingroup$ It is about $\log \log n \; . \;$ For that matter, as $n$ increases, your sum minus $\log \log n$ approaches a constant. $\endgroup$ – Will Jagy Jun 2 '18 at 1:47
  • $\begingroup$ That is, your sum (beginning at n=2) is asymptotic to $0.794 + \log \log n $ $\endgroup$ – Will Jagy Jun 2 '18 at 1:54
  • $\begingroup$ I did find that in the previous question but was confused since $\log(\log(n)) < \log(n)$, $\frac {1}{p_n} < \frac {1}{n \log(n)}$, and I also found on multiple places that $\sum_{x=1}^N \frac{1}{x} = \log(N) + O(1)$ which is similar. I think I asked my question wrong but still think it is valid. I was wondering if the previous summation could be represented as $\log(N) + f(x) + C$ where $f(x)$ is some function if possible. $\endgroup$ – jaylooker Jun 2 '18 at 16:47
  • $\begingroup$ I don't know how you concluded that your sum is larger than $\log n \; .$ The integral test for series convergence amounts to this: the sum up to $n$ and the integral up to $n$ of the same thing, when you take the difference, approaches a constant. The antiderivative of $\frac{1}{x \log x}$ is just $\log \log x$ plus a constant. $\endgroup$ – Will Jagy Jun 2 '18 at 16:52
  • $\begingroup$ I think my problem was I was thinking $\sum_{x=1}^N \frac {1}{p_x}$ ~ $\log(N)$ when it is more precisely ~ $\log(\log(n))$ as suggested here. $\endgroup$ – jaylooker Jun 2 '18 at 19:25
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First, note that $\log 1 = 0$ hence as written your sum is infinite. So let us instead consider the sum with lower index at $2$, i.e. $$S(n) = \sum_{x=2}^n \frac{1}{x \log x}.$$ Since $f(x) = 1/(x \log x)$ is convex on $x > 1$, we find that $$S(n) > \int_{x=2}^n \frac{dx}{x \log x} = \log \left(\frac{\log n}{\log 2}\right).$$ This lower bound can be improved using the trapezoidal rule, namely $$S(n) > \frac{1}{2}\log \left(\frac{\log n \log (n+1)}{\log 2 \log 3}\right).$$ Since the error is largest for small $n$, we could do still better by summing the first few terms exactly, then using the relationship $$\sum_{x=m}^n \frac{1}{x \log x} > \log \left(\frac{\log n}{\log m}\right), \quad n > m \gg 1.$$

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