Can you generalize this alternating series for $\log(2)$ to any logarithm series of a similar form?

Question

Look at the decimal expansion for the alternating sum for $\log(2)$ summed upto $n=100000$ and subtract it with the decimal expansion for the actual value for $\log(2)$:

$$\log 2=\sum_{n=1}^{100000}\frac{(-1)^{n-1}}n$$

(* Mathematica start*)
digits=100;
N[Sum[(-1)^(n + 1)/n, {n, 1, 100000}], digits]
N[Log[2], digits]
%% - %
(*end*)

Decimal expansion for the alternating sum for $\log(2)$ truncated at $n=100000$:

0.69314218058494530941598212145842656807539388436033275412059363449353\
01469694042252941638997436445553

Decimal expansion of actual value for $\log(2)$

0.69314718055994530941723212145817656807550013436025525412068000949339\
36219696947156058633269964186875

The difference between the truncated series and actual value for $\log(2)$:

-4.9999750000000012499999997500000001062499999225000000863749998634750\
00290490311699427252774132*10^-6 

Notice how it is correct at many decimal places beyond the first error.

I find that strange. It is like it suggests that there is a series acceleration that would correct this.

By guessing and adding fractions that I looked up in the OEIS, it suggests that the alternating sum together with the correcting part for parameters $p = 4$ and $r = 10$ is:

$$\log(2)=\sum _{n=1}^{6\ 10 p r} \frac{2 (-1)^n \left(\left(1-2^{-2 n}\right) \zeta (2 n) \Gamma (2 n)\right)}{\pi ^{2 n} \left(r^{2 p}\right)^n}+\sum _{n=1}^{r^p} \frac{(-1)^{n+1}}{n}+\frac{r^{-p}}{2} \;\;\;\;\;\;\;\;(\ast)$$

For $r^p=10^4=10000$, this gives the decimal expansion of $\log(2)$ to $6004$ correct digits, as shown by this Mathematica program:

(*start*)
p = 4;
r = 10;
digits = r^p;
N[Sum[(-1)^(n + 1)/n, {n, 1, r^p}] + 1/2*r^(-p) + 
   Sum[(-1)^(n)*2*(Gamma[2 n]*(1 - 2^(-2 n))*Zeta[2 n])/
       Pi^(2 n)/(r^(2*p))^n, {n, 1, 6*10*r*p}], digits];
N[Log[2], digits];
%% - %
(*end*)

If this is a known result I apologize.

What I really would like to know is how does this generalizes to series of similar form for $\log(3),\log(4),\log(5)$ and so on. Also, the requirement that $r=10$ is not optimal, and I don't know how to generalize it away so that $r^p$ can be any integer.

For $\log(3)$ with Mathematica:

I found the following repeating decimal expansion for the difference between the decimal expansion from the series:

$$\text{log3}=\sum _{n=0}^{333333} \left(\frac{1}{(3 n+1)}+\frac{1}{(3 n+2)}-\frac{2}{(3 n+3)}\right) \;\;\;\;\; (\ast\ast)$$

truncated at $n=333333$ and the actual decimal expansion for $\log(3)$:

(*start*)
nn = 333333;
s = 1;
N[Sum[1/(3*n + 1)^s + 1/(3*n + 2)^s - 2/(3*n + 3)^s, {n, 0, nn}], 100]
N[Log[3], 100]
%% - %
(*end*)

Truncated series for $\log(3)$:

1.09861128867077635139526057022385909953625944698030440640216033343750\
3992932594561746334595530692166

Decimal expansion for the actual value of $\log(3)$:

1.09861228866810969139524523692252570464749055782274945173469433363749\
4293218608966873615754813732089

Difference between decimal expansions for truncated series and actual $\log(3)$:

-9.9999733333999998466669866660511123111084244504533253400019999030028\
6014405127281159283039922*10^-7

Here again we could try to guess the fractions that are found inbetween the repeating digits:

(*Mathematica start*)
-9.9999733333999998466669866660511123111084244504533253400019999030028\
6014405127281159283039922419278832912630462820405230189615775`93.\
65812459130527*^-7
(%*3) - 4/5*10^(-6 - 5)
% + 1/5*10^(-16)
% - (1 - 54/100)*10^(-22)
% + (1 - 5/100)*10^(-28)
(*end*)
{-4/510^(-6 - 5), +1/
   510^(-16), -(1 - 54/100)10^(-22), +(1 - 5/100)10^(-28)}

The guess simplifies to the first few fractions:

$$\left\{-\frac{1}{125000000000},\frac{1}{50000000000000000},-\frac{23}{500000000000000000000000},\frac{19}{200000000000000000000000000000}\right\}$$

An OEIS search.

Question:

Can you generalize the formula in $(\ast)$ for $\log(2)$ to apply to the formula for $\log(3)$ in $(\ast\ast)$?

That is can you generalize this formula:

$$\log(2)=\sum _{n=1}^{6\ 10 p r} \frac{2 (-1)^n \left(\left(1-2^{-2 n}\right) \zeta (2 n) \Gamma (2 n)\right)}{\pi ^{2 n} \left(r^{2 p}\right)^n}+\sum _{n=1}^{r^p} \frac{(-1)^{n+1}}{n}+\frac{r^{-p}}{2} \;\;\;\;\;\;\;\;(\ast)$$

to apply to this formula: $$\log(3)=\sum _{n=0}^{\infty} \left(\frac{1}{(3 n+1)}+\frac{1}{(3 n+2)}-\frac{2}{(3 n+3)}\right) \;\;\;\;\; (\ast\ast)$$

Or even:

$$\log(4)=\sum _{n=0}^{n=\infty} \left(\frac{1}{(4 n+1)}+\frac{1}{(4 n+2)}+\frac{1}{(4 n+3)}-\frac{3}{(4 n+4)}\right)$$

and so on.

Answer 1

This is not an answer, only did the series part. Is there any idea on the series acceleration part with Zeta function?

$$\begin{align} S_n&=\sum _{i=0}^{n} \left(\frac{1}{ki+1}+\frac{1}{ki+2}+...+\frac{1}{ki+k-1}-\frac{k-1}{ki+k}\right)\\ \\ &=\sum _{i=0}^{n} \left(\frac{1}{ki+1}+\frac{1}{ki+2}+...+\frac{1}{ki+k-1}+\frac{1}{ki+k}-\frac{k}{ki+k}\right)\\ \\ &=\sum _{i=0}^{n} \left(\frac{1}{ki+1}+\frac{1}{ki+2}+...+\frac{1}{ki+k-1}+\frac{1}{ki+k}\right)-\sum _{i=0}^{n}\frac{k}{ki+k}\\ \\ &=H_{nk+k}-\sum _{i=0}^{n}\frac{1}{i+1}\\ \\ &=H_{nk+k}-H_{n+1}\\ \\ \end{align}$$

Asymptotically, $H_p=\gamma+\ln(p)+O\left(\frac{1}{p}\right)$

$$S=\lim_{n\to\infty}S_n=\lim_{n\to\infty}(H_{nk+k}-H_{n+1})=\ln(k)$$

Therefore,

$$\ln(k)=\sum _{i=0}^{\infty} \left(\frac{1}{ki+1}+\frac{1}{ki+2}+...+\frac{1}{ki+k-1}-\frac{k-1}{ki+k}\right)$$