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I would like to evaluate $$\sum_{n=2}^{\infty} \frac{1}{n\log_{2}^{2}\left(n\right)}. $$ I know that $\int_2^{\infty} \frac{1}{n\log_2^2{n}} = \log{2}$ so my guess is that the answer is close to that. However numerically it seems to be very close to $1$. Is it in fact $1$?

Using maple limit(sum(1/(n*log[2](n)^2),n=2..i), i=infinity); gives

$$\frac{48 \ln^2{2} - 48 O(1) \ln{2} + 13\ln{2} + 2}{48 \ln{2}}$$

however I have no idea how to interpret this.

Felix Marin
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user115998
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    You can approximate the value of the series using the integral. – Mhenni Benghorbal Jan 06 '14 at 20:04
  • @MhenniBenghorbal Can you tell if the answer is $1$ or not that way? – user115998 Jan 06 '14 at 20:07
  • Numerically, it is close to $1$. – Mhenni Benghorbal Jan 06 '14 at 20:14
  • But larger than 1. –  Jan 06 '14 at 20:15
  • @Karene How can you prove that? – user115998 Jan 06 '14 at 21:00
  • I don't know. I only saw an approximation that was larger than $1$. I would have to see if they approximated from below. The only thing I am sure is that if adding enough terms brings you above one then the sum is going to be larger than one because the terms are positive. –  Jan 06 '14 at 21:14
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    In Mathematica: N[Sum[1/(n (Log2[n])^2), {n, 2, Infinity}], 100] gives 1.013632287446493721670956149515565419000440623501242905035619074728115797811734508671796335269275784 –  Jan 06 '14 at 21:27
  • Imprecise. (I wonder whether something's got fixed nowadays.) I'm getting $$1.0136322874464937216709561\color{DarkBlue}{515024597538338689980226450429336228\cdots}$$ – metamorphy Feb 18 '21 at 13:52

2 Answers2

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It's straightforward to show that

$${1\over N(\ln N)^2}+{1\over(N+1)(\ln (N+1))^2}+{1\over(N+2)(\ln(N+2))^2}+\cdots\gt{1\over\ln N}+{1\over2N(\ln N)^2}$$

One thus obtains

$$\sum_{n=2}^\infty{1\over n\log_2^2n}=(\ln2)^2\sum_{n=2}^\infty{1\over n(\ln n)^2}\gt (\ln2)^2\left({1\over2(\ln2)^2}+{1\over\ln3}+{1\over6(\ln3)^2} \right)\approx1.00367$$

The "straightforward" inequality comes from comparing the series, thought of as a step function, to the integral beneath the curve $f(x)=1/x(\ln x)^2$: The difference between the two sides is the area between the curve and the chords that connect points with integer $x$ values. (The total area above those chords is $1/2N(\ln N)^2$.)

Barry Cipra
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It's possible to rewrite this sum as an integral involving the Riemann zeta function. Note that $$ \zeta(z)-1=\sum_{n=2}^{\infty}n^{-z}=\sum_{n=2}^{\infty}\exp(-z\log n) $$ when $\text{Re}(z)>1$. Taking a single antiderivative (and treating convergence casually), $$ \int_{z}^{\infty}(\zeta(s)-1)ds=\sum_{n=2}^{\infty}\int_{z}^{\infty}\exp(-s\log n)ds=\sum_{n=2}^{\infty}\frac{\exp(-z\log n)}{\log n}. $$ Taking another one, $$ \int_{z}^{\infty}dy \int_{y}^{\infty}(\zeta(s)-1)ds=\sum_{n=2}^{\infty}\int_{z}^{\infty}\frac{\exp(-y\log n)}{\log n}dy=\sum_{n=2}^{\infty}\frac{\exp(-z\log n)}{\log^2 n}, $$ or $$ \sum_{n=2}^{\infty}\frac{n^{-z}}{\log^2 n}=\int_{z}^{\infty}dy\int_{y}^{\infty}(\zeta(s)-1)ds=\int_{z}^{\infty}(\zeta(s)-1)(s-z)ds. $$ Here we want to evaluate this at $z=1$, and multiply by the correct factor to get the logarithm in base-$2$: $$ \sum_{n=2}^{\infty}\frac{1}{n\log_2^2 n}=\log^2 2\int_{1}^{\infty}(\zeta(s)-1)(s-1)ds. $$

mjqxxxx
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  • Have not checked the computation, but +1 for creativity :) –  Jan 07 '14 at 03:12
  • Thank you! But.. how does it help? :) I can't do that integral either. – user115998 Jan 07 '14 at 09:00
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    An additional change of variable gives $\log^2 2\int_{0}^{1}(\zeta(1/s)-1)(1-s)/s^3 ds$, which is at least a nice smooth function over a finite interval, and easy to approximate. – mjqxxxx Jan 07 '14 at 16:52