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There is this well known limit:

$$\lim_n \sum_{k=1}^n \frac 1k -\log n=\gamma$$

Where $\log$ is the natural logarithm and $\gamma$ is Euler constant.

I was wondering if my similar situation yelds to a similar result:

$$\lim_n \sum_{k=2}^n \frac 1{k\log k}-\log \log n=?$$

I know there is a formula (always due to Euler) which can be used in those situations but I can't see if there is a way to put the result in "closed form" rather than having only an approximation (supposing the limit exists and is finite at first)

uniquesolution
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AlienRem
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  • @Arthur I fear that too, any idea on how to prove it? – AlienRem Sep 25 '17 at 13:54
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    $\int_2^\infty (\frac{1}{\lfloor x \rfloor \log \lfloor x \rfloor}- \frac{1}{x \log x})dx$ does converge @Arthur – reuns Sep 25 '17 at 14:04
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    A quick Mathematica computation: The value for $f(n) = \sum_{k=2}^n 1/(k\ln k) - \ln\ln n$ with $n = 10'000$ to $100'000$ with step of $10'000$. ${0.794684,0.794681,0.79468,0.79468,0.79468,0.794679,0.794679,0.794679,0.794679,0.794679 }$. Looks like convergence to me. – Zubzub Sep 25 '17 at 14:12
  • Nice, I always fear the worst but I guess this is not the case! – AlienRem Sep 25 '17 at 14:16
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    The convergence can be derived from the fact the series is decreasing and a bound which can be found by comparing to integrals. (The decreasing part might prove tedious though). However, I doubt there is any closed form expression for the limit, given that there is none for the classical limit $\lim \sum \frac 1k - \log(n)$ (this is just the definition of $\gamma$). – Mariuslp Sep 25 '17 at 14:33
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    I assume the OP was referring to the Euler-Maclaurin formula, which quickly proves this converges with bounds. – Simply Beautiful Art Sep 25 '17 at 14:38
  • $$\lim_{t\to \infty } , \left(\int_2^t \frac{1}{\left(x-\frac{1}{2}\right) \log \left(x-\frac{1}{2}\right)} , dx-\log (\log (t))\right)=-\log \left(\log \left(\frac{3}{2}\right)\right)$$ can prove that the OP limit exists and is finite? – Raffaele Sep 25 '17 at 14:47
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    In this answer, this constant is evaluated to $49$ places. – robjohn Sep 25 '17 at 16:37

2 Answers2

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Using $\log\left(1+\frac1x\right)\ge\frac1{x+1}$, we get $$ \begin{align} &\left(\sum_{k=2}^{n+1}\frac1{k\log(k)}-\log(\log(n+1))\right)-\left(\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))\right)\\ &=\frac1{(n+1)\log(n+1)}-\log\left(\frac{\log(n+1)}{\log(n)}\right)\\ &\le\frac1{(n+1)\log(n+1)}-\log\left(\frac{\log(n)+\frac1{n+1}}{\log(n)}\right)\\ &=\frac1{(n+1)\log(n+1)}-\log\left(1+\frac1{(n+1)\log(n)}\right)\\ &\le\frac1{(n+1)\log(n+1)}-\frac1{(n+1)\log(n)+1}\\ &\le\frac1{(n+1)\left(\log(n)+\frac1{n+1}\right)}-\frac1{(n+1)\log(n)+1}\\ &=0 \end{align} $$ Therefore, $$ \sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n)) $$ is decreasing. Furthermore, $$ \begin{align} \sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n)) &\ge\int_2^{n+1}\frac{\mathrm{d}x}{x\log(x)}-\log(\log(n))\\ &=\log(\log(n+1))-\log(\log(2))-\log(\log(n))\\[9pt] &\ge-\log(\log(2)) \end{align} $$ is bounded below,

Thus, $$ \lim_{n\to\infty}\left(\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))\right) $$ exists.

In fact, in this answer, this limit is computed to $49$ places using the Euler-Maclaurin Sum Formula: $$ 0.7946786454528994022038979620651495140649995908828 $$

robjohn
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Bit long for a comment, not really a finished answer.

More generally, consider

$$I(s)=\int_2^\infty\left(\frac1{\lfloor x\rfloor^s\ln\lfloor x\rfloor}-\frac1{x^s\ln(x)}\right)~\mathrm dx$$

By differentiating under the integral,

$$I'(s)=\int_2^\infty\left(\frac1{\lfloor x\rfloor^s}-\frac1{x^s}\right)~\mathrm dx$$

Which may be shown to equal

$$I'(s)=\begin{cases}\zeta(s)-1+\frac1{1-s}2^{1-s},&s\ne1\\\gamma,&s=1\end{cases}$$

And so,

\begin{align}I(1)&=\int_1^\infty\left(1-\zeta(s)-\frac{2^{1-s}}{1-s}\right)~\mathrm ds\\&=\gamma-1+\lim_{b\to\infty}\big(b-\zeta^{\star}(b)\big)-\lim_{a\to0^+}\big(\ln(-a\ln(2))-\zeta^\star(a+1)\big)\end{align}

where

$$\zeta^\star(s)=\int_t^s\zeta(x)~\mathrm dx,\quad t>1$$

and $\operatorname{Ei}(s)$ is the exponential integral.

...not really sure if this integral/limit is doable...

Simply Beautiful Art
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