High precision evaluation of the series $\sum_{n=3}^\infty (-1)^n (1-n^{1/n})$

Question

This series converges conditionally, but it's quite slow. I would like to find its value with high accuracy:

$$S=\sum_{n=3}^\infty (-1)^n (1-n^{1/n})$$

Wolfram Alpha gives $S \approx 0.226354\ldots$.

Since the terms decrease monotonely in absolute value, we can apply an approximate estimation:

$$S_N= \sum_{n=3}^{N-1} (-1)^n (1-n^{1/n})+ \frac{1}{2} (-1)^N (1-N^{1/N})$$

$$S_{100}=0.22644\ldots$$

$$S_{101}=0.22626\ldots$$

$$\frac{S_{100}+S_{101}}{2} =0.22635473854439942\ldots$$

Another way could be to transform the series, for example:

$$n^{1/n}=\exp \frac{\log n}{n}=\sum_{k=0}^\infty \frac{\log^k n}{n^k k!}$$

Which gives us (assuming we are allowed to change the order of summation):

$$S=\sum_{k=1}^\infty \frac{1}{k!} \sum_{n=3}^\infty (-1)^{n+1} \frac{\log^k n}{n^k}=\sum_{k=1}^\infty \frac{S_k}{k!}$$

The inner series $S_k>0$ can be expressed in terms of repeated derivatives of the zeta function, which don't have a closed form for $k \geq 2$, but the series can still be evaluated numerically with high accuracy.

---

Note

$$S_1=\frac{\log 2}{2} (1+\log 2-2\gamma)$$

For $k \geq 2$ we can easily write:

$$S_k=\sum_{q=1}^\infty \frac{\log^k (2q+1)}{(2q+1)^k}-\sum_{q=2}^\infty \frac{\log^k (2q)}{(2q)^k}$$

Both the series converge absolutely and can be easily approximated by Euler-Maclaurin summation with all the integrals and derivatives expressed in closed form (obviously for large $k$ it becomes unwieldy).

---

Evaluating the series up to $S_6$ we obtain:

$$S > 0.2263538 \ldots$$

Still not that good.

Finally, we could use Euler-Maclaurin, but I'm not sure how to apply it in this case, especially how to deal with the integral.

Answer 1

Using PARI/GP you can get a thousand decimal digits using

default(realprecision,10^3); c=sumalt(n=3, (-1)^n*(1-n^(1./n)));

in under 30 seconds. It takes much longer for more digits. It Seems to be an $O(n^3)$ time algorithm where $n$ is the number of digits.

In case you are interested, the documentation for PARI/GP describes the algorithms used for the summation of alternating series. Also, the entire source code is under the GPL license. Here is a brief documentation excerpt.

sumalt(X = a,expr,{flag = 0}):

Numerical summation of the series expr, which should be an alternating series $(-1)^k a_k,\,$ the formal variable X starting at a. Use an algorithm of Cohen, Villegas and Zagier (Experiment. Math. 9 (2000), no. 1, 3--12).

The work mentioned here seems to be available from a Project Euclid link. You may also be interested in the OEIS sequence A037077

Decimal expansion of upper limit of - 1^(1/1) + 2^(1/2) - 3^(1/3) + ...

authored by Marvin Ray Burns for which the constant is named (MRB constant).

Answer 2

Hints.

$\displaystyle \sum\limits_{n=3}^\infty (-1)^{n-1}(\sqrt[n]{n}-1) = \sqrt{2}-1 + \sum\limits_{k=1}^\infty \frac{1}{k!} \sum\limits_{n=1}^\infty (-1)^{n-1}\left(\frac{\ln n}{n}\right)^k $

$\displaystyle \sum\limits_{n=1}^\infty (-1)^{n-1}\left(\frac{\ln n}{n}\right)^k |_{k=1} = \frac{\ln 2}{2}(\ln 2~-2\gamma) $

$\displaystyle \sum\limits_{n=1}^\infty (-1)^{n-1}\left(\frac{\ln n}{n}\right)^k |_{k=2} = \frac{1}{2}\zeta’’(2) + \frac{\pi^2\ln 2}{12}(\ln 2 + 2\gamma + 2 \ln\pi~ – 24\ln A)$

$\hspace{7cm}$ where $A$ is the Glaisher-Kinkelin constant

I have no proof for the following formula:

$\displaystyle \sum\limits_{n=1}^\infty (-1)^{n-1}\left(\frac{\ln n}{n}\right)^k |_{k\geq 3} = \sum\limits_{j=0}^k a_{k,j}(\ln 2)^{k-j}\zeta^{(j)}(k)$

with $\enspace\displaystyle a_{k,j}|_{0\leq j\leq k-1}=\frac{(-1)^{j+1}}{2^{k-1}}\binom k j\enspace$ and $\enspace\displaystyle a_{k,k} = (-1)^k\frac{2^{k-1}-1}{2^{k-1}}$

Informations about $~\zeta^{(j)}(k)~$ please see e.g. here , formulas (11)+(16)+(19).