Why $\sum _ {n = x}^{\infty} (-1)^n\left (n^{1/n} - 1 \right) $ gives so many repeating 0 s here?

Question

$\sum _ {n = 1}^{\infty} (-1)^n\left (n^{1/n} - 1 \right) $ is the MRB constant.

Whether it's rational or not I know its terms and all of its partial sums are irrational.

So why does $\sum _ {n = x}^{\infty} (-1)^n\left (n^{1/n} - 1 \right) $ gives so many repeating 0 s here?

Specifically, I found the following that I would like some insight into.

Using Mathematica,

f[x_] :=NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, Method -> "AlternatingSigns", WorkingPrecision ->200]

 Table[N[(10 f[10^(10 n)])/Log[10], 10 n + 5], {n, 1, 10}] // TableForm

gives the following:

{"5.00000000599560537540642761618*^-9"}, {"1.0000000000000000002351499356789287554825*^-18"}, {"1.5000000000000000000000000000525473072303184465947*^-28"}, {"2.0000000000000000000000000000000000000093092546357714246065*^-38"}, {"2.5000000000000000000000000000000000000000000000014515071095008027395*^-48"}, {"3.00000000000000000000000000000000000000000000000000000000020872180100741653026*^-58"}, {"3.5000000000000000000000000000000000000000000000000000000000000000000283805816529723015*^-68"}, {"4.00000000000000000000000000000000000000000000000000000000000000000000000000000370402757516999728*^-78"}, {"4.50000000000000000000000000000000000000000000000000000000000000000000000000000000000000046851262396924667*^-88"}, {"5.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000057813541588646383970880658544811814*^-98"}

I know that $$x^{1/x} = e^{\frac {\log x} {x}},$$

but I still don't see where all the repeating 0s come from.

Answer 1

This is what you would expect if $$\sum _ {n = x}^{\infty} (-1)^n\left (n^{1/n} - 1 \right) \approx \frac12 (-1)^x \frac{\log_e(x)}{x}$$ with the error being much smaller - perhaps of the order of the square of that.

For example if $x=10^{10}$ (your first example) this would suggest about $\frac 12 10 \log_e(10) \times 10^{-10}\approx 1.151293\times 10^{-9}$ but you divide this by $\log_e(10)$ so getting $5 \times 10^{-9}$ with an error of around $5.9956 \times 10^{-18}$ so getting several zeros.

Meanwhile with $x=10^{100}$ (your last example) this would suggest about $\frac 12 100 \log_e(10) \times 10^{-100}$ but you divide this by $\log_e(10)$ so getting $5 \times 10^{-98}$ with an error of around $5.78 \times 10^{-196}$ so getting many more zeros.