This is similar to my recent question, but probably more interesting. What can we say about this slowly converging series:
$$S=\sum_{n=2}^\infty \left(n^{1/n}-1-\frac{\log n}{n} \right)$$
We can rewrite it as a double sum:
$$S=\sum_{n=2}^\infty \sum_{k=2}^\infty \frac{\log^k n}{k!~n^k}$$
Or express through zeta derivatives at integer points:
$$S=\sum_{k=2}^\infty \frac{| \zeta^{(k)}(k) |}{k!}$$
Using the integral expression for zeta derivatives I obtained yesterday* and exchanging integration and summation, we can claim:
$$S=\sum_{k=2}^\infty \frac{1}{(k-1)^{k+1}}+$$
$$+ \frac{1}{i} \int_0^{\infty } \frac{dt}{\exp (2 \pi t)-1} \left((1-i t)^{\frac{1}{1-i t}}-(1+i t)^{\frac{1}{1+i t}}-\frac{\log (1-i t)}{1-i t}+\frac{\log (1+i t)}{1+i t} \right)$$
This gives us the following numerical value:
$$S=1.0613654466259454755053404842148725835597350221360 \ldots$$
Directly evaluating the series in Mathematica is hard, but it gives an approximately the same value $1.06136544662594547550535127514661 \ldots$.
Questions:
- Is my numerical value correct? What other methods can we use to check? Euler-Maclaurin doesn't seem to work well, mostly because the integral is not nice eihter.
(The original version of this question contained a mistake, the corrected integral expression gives the same value as numerical summation for the first several digits, but I would still like to know more correct digits).
- Is there a way to obtain a nicer integral expression for the series?
$*$ $$\zeta^{(n)}(s)=\frac{(-1)^n n!}{(s-1)^{n+1}}- \\ +\frac{1}{i} \int_0^{\infty } \frac{dt}{e^{2 \pi t}-1} \left(\frac{(1+i t)^s}{\left(1+t^2\right)^s} \log ^n\left(\frac{1+i t}{1+t^2}\right)-\frac{(1-i t)^s }{\left(1+t^2\right)^s}\log ^n\left(\frac{1-i t}{1+t^2}\right) \right)$$