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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.

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    $\begingroup$ You can approximate the value of the series using the integral. $\endgroup$ – Mhenni Benghorbal Jan 6 '14 at 20:04
  • $\begingroup$ @MhenniBenghorbal Can you tell if the answer is $1$ or not that way? $\endgroup$ – user115998 Jan 6 '14 at 20:07
  • $\begingroup$ Numerically, it is close to $1$. $\endgroup$ – Mhenni Benghorbal Jan 6 '14 at 20:14
  • $\begingroup$ But larger than 1. $\endgroup$ – user119256 Jan 6 '14 at 20:15
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    $\begingroup$ In Mathematica: N[Sum[1/(n (Log2[n])^2), {n, 2, Infinity}], 100] gives 1.013632287446493721670956149515565419000440623501242905035619074728115797811734508671796335269275784 $\endgroup$ – user114628 Jan 6 '14 at 21:27
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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 $1/2N(\ln N)^2$.)

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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-$2$: $$ \sum_{n=2}^{\infty}\frac{1}{n\log_2^2 n}=\log^2 2\int_{1}^{\infty}(\zeta(s)-1)(s-1)ds. $$

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  • $\begingroup$ Have not checked the computation, but +1 for creativity :) $\endgroup$ – user114628 Jan 7 '14 at 3:12
  • $\begingroup$ Thank you! But.. how does it help? :) I can't do that integral either. $\endgroup$ – user115998 Jan 7 '14 at 9:00
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    $\begingroup$ 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. $\endgroup$ – mjqxxxx Jan 7 '14 at 16:52

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