How to compute this constant with high precision $\sum_{n=1}^\infty \left(\frac{1}{a_n}-\frac{1}{(n+1) \ln (n+1)} \right)$

Question

I'm interested in finding the following constant:

$$b=\sum_{n=1}^\infty \left(\frac{1}{a_n}-\frac{1}{(n+1) \ln (n+1)} \right)$$

Where:

$$a_1=2$$

$$a_{n+1}=a_n+\log a_n$$

This is related to my recent question where the sequence was first introduced and it was shown in the answer that:

$$\lim_{n \to \infty} \frac{a_n}{n \ln n}=1$$

I wanted to see what the constant above looks like, because this is similar to how the Euler-Mascheroni constant is obtained from the harmonic series and the logarithm.

The problem is, the convergence of the above series is extremely slow. And I mean so slow, that I'm not even sure what the first digit is.

From Mathematica computations it seems that:

$$0.1 <b <0.2$$

But I'm only sure about the upper bound, because $b$ becomes smaller as the number of terms increases.

Note that even though initial partial sums are negative, $b$ becomes positive soon, because the sequence $a_n$ gets overtaken by $(n+1) \ln (n+1)$, even if they are of the same order. You can see that in the linked question.

Then $a_n$ overtakes $(n+1) \ln (n+1)$ again, and partial sums start to decrease.

Mathematica gives:

$$\sum_{n=1}^{10^7} \left(\frac{1}{a_n}-\frac{1}{(n+1) \ln (n+1)} \right)=0.18702446577 \dots$$

But at least the second digit is different from the true value of $b$, as can be seen by adding further terms.

$$\sum_{n=1}^{10^8} \left(\frac{1}{a_n}-\frac{1}{(n+1) \ln (n+1)} \right)=0.1738163796928 \dots$$

(In case it's important, I was keeping only $100$ digits of each $a_n$ while computing the recurrence terms. Maybe there's some loss of precision there as well).

Update

$$\sum_{n=1}^{10^9} \left(\frac{1}{a_n}-\frac{1}{(n+1) \ln (n+1)} \right)=0.162 \dots$$

(I have more digits, but it's clear they don't matter at this point).

And for some $10^9<N<10^{10}$ (aborted computation) we have:

$$b<0.1599565$$

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Can we find at least a few first digits of $b$? What methods would you suggest for accelerating the series or transforming it somehow for faster convergence?

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The motivation for this question is not the sequence itself (I don't believe it has any significance) but rather methods for solving this kind of problems.

As a side question, can we at least prove that the series converges? I'm rather convinced it does, but just in case.

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For other series related to the sequence we have, reliably:

$$\sum_{n=1}^\infty \frac{1}{a_n^2}=0.57409540\dots$$

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{a_n}=0.285408\dots$$

The convergence is quite slow for these cases as well, which was expected by comparison with known series.

Answer 1

$\DeclareMathOperator{\li}{li}$ I can confirm the convergence of the series, but it is the slowest convergence I have ever encoutered in a series not constructed for this purpose. Here the approximation of $a_n$ using the logarithmic integral function is used. The tail of the series can be evaluated using the Euler-MacLaurin formula and the value $0.02225375...$ is found using the first 100 terms of the series and evaluation of the tail. All 8 decimals are correct. The approximation of $a_n$ is again crucial as well as the fact that the integrals occuring can be expressed using elementary functions.

For a proof of the convergence, we only need that the sequence $a_n$ defined by $a_1=2$ and $a_{n+1}=a_n+\log(a_n)$ satisfies $a_n=\li^{-1}(n)+O(\log(n)^2)=(n+1)(\log(n+1)+O(\log(\log(n))))$ which follows form this answer. It implies that $$\frac1{a_n}-\frac1{(n+1)\log(n+1)}= O\left(\frac{\log(\log(n))}{n\log(n)^2}\right).$$ Now, as shown in Yurij S' answer, $\sum_{n=3}^\infty\frac{\log(\log(n))}{n\log(n)^2} $ converges because the corresponding integral $\int_{3}^\infty\frac{\log(\log(t))}{t\log(t)^2}\,dt=\int_{\log(3)}^\infty\frac{\log(s)}{s^2}\,ds$ converges. This proves the convergence of the series. It also permits to show the slow convergence: If we sum the tail of the above series from $n=N$ to infinity then the integral is of the order $\log(\log(N))/\log(N)$. By summing $10^8$ terms, the tail of the given series then still is (a multiple of) $0.158...$. It is hopeless to try to evaluate the series by summing its terms even with a very fast computer.

To evaluate the given series, we use in a first step the approximation of this answer. There is a constant $C$ such that $$g(a_n)=n+C+O\left(\frac1{n\log(n)}\right),\mbox{ where } g(x)=\li(x)+\frac12\log(\log(x)).$$ Here $\li$ denotes the logarithmic integral function (see this page.) As $$g'(x)=\frac1{\log(x)}+\frac1{2x\log(x)},$$ we have in particular $(g^{-1})'(y)=\frac1{g'(g^{-1}(y))}\sim \log(g^{-1}(y))\sim \log(y)$ and hence $a_n=g^{-1}(n+C)+O\left(\frac1{n}\right)$. This yields $$\frac1{a_n}-\frac1{g^{-1}(n+C)}=O\left(\frac1{n^3\log(n)^2}\right).$$ Hence for $M>N$ sufficiently large $$\sum_{n=N}^M\left(\frac1{a_n}-\frac1{g^{-1}(n+C)}\right)=O\left(\frac1{N^2\log(N)^2}\right).$$

$\newcommand{\ds}{\displaystyle}$ In a secong step, we apply the Euler-MacLaurin formula with $h(t)=1/g^{-1}(t+C)$: $$\begin{array}{rcl}\ds\sum_{n=N}^M h(n)&=&\ds\int_N^M h(t)\,dt+\frac12h(N)+\frac12h(M)+\frac1{12}h'(M)-\frac1{12}h'(N) +O\left(\int_N^M|h^{(2)}(t)|\,dt\right)\\&=&\ds\int_N^M h(t)\,dt+\frac12h(N)+\frac12h(M)+O\left(\frac1{N^2\log(N)}\right).\end{array}$$ The proof of the estimate for the error terms is omitted. We now evaluate each of the terms in the above formula. Using the substitution $t=g(s)-C$, we find $$\begin{array}{ccl}\ds\int_N^M h(t)\,dt&=&\ds\int_N^M \frac1{g^{-1}(t+C)}\,dt= \ds\int_{g^{-1}(N+C)}^{g^{-1}(M+C)} \frac{1}sg'(s)\,ds=\\ &=&\ds\int_{g^{-1}(N+C)}^{g^{-1}(M+C)} \left( \frac1{s\log(s)}+\frac1{2s^2\log(s)} \right)\,ds\\ &=&\ds \left( \log(\log(s))+\frac12\li(1/s) \middle)\right|_{s=g^{-1}(N+C)}^{s=g^{-1}(M+C)}=\ds \left( \log(\log(s))+\frac12\li(1/s) \middle)\right|_{a_N}^{a_M}+O\left(\frac1{N^2\log(N)}\right). \end{array}$$ In the last equality, $a_n=g^{-1}(n+C)+O\left(\frac1{n}\right)$ has been used again. This gives altogether $$\ds\sum_{n=N}^M h(n)=\ds \left( \log(\log(s))+\frac12\li(1/s) \middle)\right|_{a_N}^{a_M}+\frac1{2a_N}+\frac1{2a_M}+O\left(\frac1{N^2\log(N)}\right)$$

Similarly, but simpler since we are dealing with an elementary function, we find $$\ds\sum_{n=N}^M\frac1{(n+1)\log(n+1)}=\log(\log(s))|_{N+1}^{M+1}+\frac12\frac1{(N+1)\log(N+1)}+ \frac12\frac1{(M+1)\log(M+1)}+O\left(\frac1{N^2\log(N)}\right).$$

Now we combine the above estimates and let $M$ tend to $\infty$. Using that $ \log(\log(a_M)) - \log(\log(M+1))\to0$ since $a_M\sim M\log(M)$, we obtain the wanted approximations of the tails $$\begin{array}{rcl}\ds\sum_{n=N}^\infty\left(\frac1{a_n}-\frac1{(n+1)\log(n+1)}\right)&=& \ds- \log(\log(a_N))-\frac12\li(1/a_N)+\frac1{2a_N}+\\&&+\ds \log(\log(N+1))-\frac12\frac1{(N+1)\log(N+1)} +O\left(\frac1{N^2\log(N)}\right).\end{array}$$

Finally we can evaluate the given series by summing its first $N-1$ terms and using the above approximation of the tail. Here $N\in\mathbb N$ is a parameter. We obtain $$\begin{array}{rcl}b=\ds\sum_{n=1}^\infty\left(\frac1{a_n}-\frac1{(n+1)\log(n+1)}\right)&=& \ds\sum_{n=1}^{N-1}\left(\frac1{a_n}-\frac1{(n+1)\log(n+1)}\right)\\&& \ds- \log(\log(a_N))-\frac12\li(1/a_N)+\frac1{2a_N}\\&&+\ds \log(\log(N+1))-\frac12\frac1{(N+1)\log(N+1)}\\&&\ds +O\left(\frac1{N^2\log(N)}\right).\end{array}$$

Using this formula for $N=100$ gives the value $0.02225375...$ for the series; comparison with the results for larger $N$ show that all 8 decimals after the dot are correct. Using $N=10^7$ as in the question and a refined (more complicated) formula having an error term $O\left(\frac1{N^3\log(N)}\right)$ provides a value $0.0222\ 5375\ 6202\ 8220\ 6538\ 1016...$ with $24$ correct decimals.

Answer 2

I'll start posting my own results in this answer now. It will be updated as I progress.

Using a hint by didgogns, I have obtained numerically the following upper bound for $n \geq 2$, which works at least until $n=10^6$, and I don't have any reason to suspect it would break:

$$a_n \leq (n+1) \left( \ln (n+1) + \ln \ln (n+1) \right) \tag{1}$$

I will try proving this by induction.

The base case is $n=2$:

$$a_2=2+\ln 2=2.69314718\dots$$

$$3( \ln 3+ \ln \ln 3)=3.57798 \dots$$

The induction hypothesis is (shifting the index by $1$ for convenience):

$$a_{n-1} \leq n \left( \ln n + \ln \ln n \right)$$

Now consider $a_n$:

$$a_n=a_{n-1}+ \ln a_{n-1} \leq n \left( \ln n + \ln \ln n \right)+ \ln \left(n \left( \ln n + \ln \ln n \right) \right)$$

We try to prove that:

$$f(n) \leq g(n)$$

Where:

$$f(n)=n \left( \ln n + \ln \ln n \right)+ \ln \left(n \left( \ln n + \ln \ln n \right) \right)$$

$$g(n)=(n+1) \left( \ln (n+1) + \ln \ln (n+1) \right)$$

For $n \geq 2$ Mathematica seems to confirm this numerically quite well:

Since there's a global minimum, I could try proving the inequality by calculus methods. Some transformations give:

$$f(n)=(n+1) \ln n+(n+1) \ln \ln n+\ln \left(1 + \frac{\ln \ln n}{\ln n} \right)$$

$$g(n)=(n+1) \ln n+(n+1)\ln \left(1+ \frac{1}{n} \right) +(n+1) \ln \left( \ln n+\ln \left(1+ \frac{1}{n} \right)\right)$$

So we have:

$$g(n)-f(n)=(n+1)\ln \left(1+ \frac{1}{n} \right)+(n+1)\ln \left( 1+\frac{\ln \left(1+ \frac{1}{n} \right)} {\ln n} \right)-\ln \left(1 + \frac{\ln \ln n}{\ln n} \right)$$

It's obvious that:

$$\lim_{n \to \infty} (g(n)-f(n))=1$$

We could try using the Taylor series (first two terms) to simplify the expression and prove that it's always positive. I might do it later, but for now I will believe Mathematica.

It gives for the root of $g'-f'$ around $x=20000$ the following value:

$$x_0=12349.87656\dots \\ g(x_0)-f(x_0)=0.89262929 \dots$$

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Using the bound (1) we have for some $N$:

$$b \geq \sum_{n=1}^N \left(\frac{1}{a_n}-\frac{1}{(n+1) \ln (n+1)} \right)+ \\ +\sum_{n=N+1}^\infty \frac{1}{(n+1) } \left(\frac{1}{\ln (n+1)+\ln \ln (n+1)}-\frac{1}{\ln (n+1)} \right)$$

Or:

$$b \geq \sum_{n=1}^N \left(\frac{1}{a_n}-\frac{1}{(n+1) \ln (n+1)} \right)-\sum_{n=N+2}^\infty \frac{\ln \ln n}{n \ln n \left( \ln n+\ln \ln n \right)}$$

The series still needs to be evaluated, and Mathematica has trouble with it.

However, from the results Mathematica gives (accounting for the error it claims in the warning message) we have for $N=10^9$:

$$b>0.162\ldots-0.181\ldots=-0.019\ldots$$

So far this is the best result I've got. It seems that $b$ could be negative after all.

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Important result!

We now can prove convergence by showing that the following series converges by using the integral test:

$$\sum_{n=3}^\infty \frac{\ln \ln n}{n \ln n \left( \ln n+\ln \ln n \right)} \leq \frac{\ln \ln 3}{3 \ln 3 \left( \ln 3+\ln \ln 3 \right)}+ \int_3^\infty \frac{\ln \ln x ~ dx}{x \ln x \left( \ln x+\ln \ln x \right)}$$

The integral can be simplified:

$$\int_3^\infty \frac{\ln \ln x~ dx}{x \ln x \left( \ln x+\ln \ln x \right)}=\int_{\ln \ln 3}^\infty \frac{y ~dy}{e^y+y}=0.802467208941\dots$$

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The integral bounds actually give extremely accurate way to evaluate the series, for example:

$$\sum_{n=10^9}^\infty \frac{\ln \ln n}{n \ln n \left( \ln n+\ln \ln n \right)}=0.181044136\dots$$

Where all the digits are correct.