Closed form $\sum_{n=2}^{\infty} \frac{1}{\ln^n{n}}$ and $\sum_{n=2}^{\infty} \frac{n}{\ln^n{n}}$ Apologies if this has been asked before, but I was playing around with Wolfram Alpha and got approximations but not closed forms for 
$$\sum_{n=2}^{\infty} \frac{1}{\ln^n{n}} \approx 3.2426094109 $$
and
$$\sum_{n=2}^{\infty} \frac{n}{\ln^n{n}} \approx 8.2527103547 $$
I would like to know if a closed form exists or not.
 A: It is unlikely that a closed form exists (there are very few series that have closed forms). 
Of course, the definition of closed form is somewhat ambiguous. However, lets define it as follows:
Let $C(f)=$the number of basic arithmetic operations required to compute expression $f$.
I would say that we have a strict "closed form" expression $f$ for an infinite series $S$  iff $f_i=S_i$ for all partial sums of sequence values 1 to $i$.
We have asymptotically closed form if $C(s_i)=o(C(S_i))$.
However, for your case the associated integrals do not have a closed form, so why would you expect the infinite series to have one?
In general, each denominator involves a different value for the log and a different exponent, so there is no algebraic "shortcut" that effectively collapses the values to a nice expression. 
For example, a geometric series is nice because each term is a series of exponents with a common base. This general idea applies to even broader items like telescoping series. In general, I think that the associated sequence needs to be recursively enumerable for there to be a nice shortcut formula:
$S_i=f(S_{i-1})$ for some fixed function $f(\cdot)$ that is independent of $i$.
For your function $(\ln (n))^{-n}\neq f((\ln (n-1))^{-n+1})$ so the answer is likely no. 
