Why's $\sum _{n=1}^{\infty } (-1)^n (n^{1/n}-1)$ accelerated by $\sum _{n=1}^{\infty } \frac{1}{2} (-1)^n \left(n^{1/n}-(n+1)^{\frac{1}{n+1}}\right)?$

Question

The series I started with is defined here. The mathematical codes used below are

 m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
WorkingPrecision -> 200, Method -> "AlternatingSigns"];

Table[Print[
 N[-MantissaExponent[
   NSum[(-1)^n (n^(1/n)), {n, 1, 10^x}, WorkingPrecision -> 200, 
     Method -> "AlternatingSigns"] - m][[2]]], 
" digits come from ", "10^", x, " partial sums"], {x, 1, 51}];

and

Table[Print[
N[-MantissaExponent[
   NSum[(-1)^n (n^(1/n) - (n + 1)^(1/(n + 1))), {n, 1, 10^x}, 
      WorkingPrecision -> 200, Method -> "AlternatingSigns"]/2 - 
    m][[2]]], " digits come from ", "10^", x, 
" partial sums"], {x, 1, 51}];

and dr = Table[ NSum[(-1)^n (n^(1/n) - 1), {n, 1, x}, WorkingPrecision -> 20], {x, 1, 100}] - m;

 dr1 = Table[
NSum[((2 k)^(1/(2 k)) - (2 k - 1)^(1/(2 k - 1))), {k, 1, x}, 
 WorkingPrecision -> 20], {x, 1, 100}] - m;

 dn = Table[
 NSum[(-1)^n (n^(1/n) - (n + 1)^(1/(n + 1))), {n, 1, x}, 
  WorkingPrecision -> 20], {x, 1, 100}]/2 - m;
ListPlot[{dr, dr1, dn}, 
PlotLegends -> {(-1)^n (n^(1/n) - 1) - 
CMRB, ((2 k)^(1/(2 k)) - (2 k - 1)^(1/(2 k - 1))) - 
CMRB, (-1)^n ((n^(1/n) - (n + 1)^(1/(n + 1)))/2) - CMRB}]

I was looking for a way to improve the MRB constant's series convergence rate and found one while thinking about the trapezoidal rule for integrals. Is there a general formula for that acceleration in some analogy to the Newton-Cotes formulas?

Here is how I found it with Mathematica

P.S I noticed some kind of pattern arising, but I can't put my finger on it. Can you help?

 m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, WorkingPrecision -> 20, Method -> "AlternatingSigns"];

2 m - Sum[(-1)^n (n^(1/n) - (n + 1)^(1/(n + 1))), {n, 1,Infinity}];N[%]

 (Sqrt[2] - 1) - Sum[(-1)^n (n^(1/n) - (n + 2)^(1/(n + 2))), {n, 1, Infinity}];N[%]

2 m - (-3^(1/3) + Sqrt[2]) - Sum[(-1)^n (n^(1/n) - (n + 3)^(1/(n + 3))), {n, 1, Infinity}];N[%]

Here is the plot, with this new series, d3.

    dr = Table[
NSum[(-1)^n (n^(1/n) - 1), {n, 1, x}, WorkingPrecision -> 20], {x,
  1, 100}] - m;

  dr1 = Table[
NSum[((2 k)^(1/(2 k)) - (2 k - 1)^(1/(2 k - 1))), {k, 1, x}, 
 WorkingPrecision -> 20], {x, 1, 100}] - m;

   dn = Table[
 NSum[(-1)^n (n^(1/n) - (n + 1)^(1/(n + 1))), {n, 1, x}, 
  WorkingPrecision -> 20], {x, 1, 100}]/2 - m;

    d3 = Table[
 NSum[(-1)^n (n^(1/n) - (n + 3)^(1/(n + 3))), {n, 1, x}, 
  WorkingPrecision -> 20], {x, 1, 100}]/2 -   m + (-3^(1/3) + 
   Sqrt[2])/2;

   ListPlot[{dr, dr1, dn, d3},PlotLegends -> {"\[Sum] 
(-1)^n(n^(1/n)-1)-CMRB", 
 "\[Sum](2k)^(1/(2k))-(2k-1)^(1/(2k-1))-CMRB", 
 "\[Sum](-1)^n((n^(1/n)-(n+1)^(1/(n+1)))/2)-CMRB", 
 "\[Sum](-1)^n (n^(1/n)-(n+3)^(1/(n+3)))/2+(-3^(1/3)+Sqrt[2])/2- 
 CMRB"}]

'

@KCd lead me to ACCELERATING CONVERGENCE OF SERIES by KEITH CONRAD here.

It looks like

Answer 1

I find it hard to read the body of your post to see what it is you are focusing on (it looks like a wall of code). So what I say here may not be helpful.

This method of acceleration is called Euler’s transformation. See section 3 here. Pages 11 and 12 there describe what iterations of this acceleration method look like; see equation (3.21).