RootSum[] is necessary for reasons similar to why Root[] is needed: namely, as a way to represent sums of a function over the roots of a polynomial. In brief, if you have the expression
r == RootSum[C[0] + C[1] # + ... + C[n] #^n &, f[#] &]
this is equivalent to
$$r=\sum_{k=1}^n f(x_k),\qquad c_n x_k^n+\cdots+c_1 x_k+c_0=0,\; k=1,\dots,n$$
In the case of the expression you have, note that the generating polynomial (i.e. the first argument of RootSum[]) has the following structure:
Collect[1 + 2 #1^2 + 6 n #1^2 - 6 n^2 #1^2 + 2 n^3 #1^2 - 23 #1^4 + 50 n #1^4 - 25 n^2 #1^4 - 12 n^3 #1^4 + 15 n^4 #1^4 - 6 n^5 #1^4 + n^6 #1^4 + 36 #1^6 - 168 n #1^6 + 316 n^2 #1^6 - 304 n^3 #1^6 + 156 n^4 #1^6 - 40 n^5 #1^6 + 4 n^6 #1^6 &[x], x] 1 + (2 + 6 n - 6 n^2 + 2 n^3) x^2 + (-23 + 50 n - 25 n^2 - 12 n^3 + 15 n^4 - 6 n^5 + n^6) x^4 + (36 - 168 n + 316 n^2 - 304 n^3 + 156 n^4 - 40 n^5 + 4 n^6) x^6
and you should immediately observe that all the powers involved are even. This means that if we let $u=x^2$, we get a polynomial that is cubic in $u$, which lends itself to using things like Cardano's formula to express the roots.
I won't go that far. (The sufficiently masoch- ...er, motivated reader is free to go further than me.) Instead, the obervation of even powers is why you get the following result after applying Normal[]:
and you should now notice that you obtain a result of the form $\sum_k\frac{\log(-x_k)}{x_k}$, where the $x_k$ come in $\pm\sqrt{\cdot}$ pairs, corresponding to the roots of the cubic polynomial I mentioned earlier. You could try using ToRadicals[] to get the Cardano-like radical form, but it won't be pretty.
Good luck!
n. If you choosen=1you get $4\pi$ :FullSimplify[thething /. n -> 1]$\endgroup$nthat work e.gFullSimplify[ToRadicals[thething/.n->4]]gives you $2 \left(4+\sqrt{7}\right) \pi$,n=3gives you $8\pi$, andn=1gives you $4\pi$, but most others e.gn=6do not simplify to anything nice. $\endgroup$