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We can try to compute using bare hands. The following is thus not useful if the question wants more than this particular case. I cannot see how sage supports better this now.

First of all, we can compute $\displaystyle L(\chi,1)=\sum_{n\ge 0}\left( \frac 1{3n+1} - \frac 1 {3n+2}\right)=\sum_{n\ge 0}\frac 1{(3n+1)(3n+2)}$ exactly, for instance:

sage: sum( 1/(3*n+1)/(3*n+2), n, 0, oo )                           

and the result is connected to "polylogarithmic computations".

Comment: We may start with $1+x^3 +x^6+\dots =1/(1-x^3)$ and integrate twice, first with $x$ from $0$ to $y$, then with $y$ from $0$ to $1$. Fubini shows we can forget about polylogarithms, since $$ L(\chi, 1) = \int_0^1 dy\int_0^ydx\; \frac 1{1-x^3} = \int_0^1dx\int_x^1dy\; \frac 1{1-x^3} =\int_0^1\frac {dx}{1+x+x^2}=\frac 1{\sqrt 3}\pi\ .$$

The derivative is more complex. Pari/GP gave

? sum( n=0,10000000, -log(3*n+1)/(3*n+1)+log(3*n+2)/(3*n+2) )
%6 = 0.2226631782653383756620209560

but suminf( n=0,-log(3*n+1)/(3*n+1)+log(3*n+2)/(3*n+2) ) was testing my patience.

At any rate, the rest can be estimated by rewriting the sum as a sum over $$ \frac 1{(3n+1)(3n+2)} \left[\ \ln \left(1+\frac 1{3n+1}\right)^{3n+2} -\ln(3n+2) \ \right] $$ or over $$ \frac 1{(3n+1)(3n+2)} \left[\ \ln \left(1+\frac 1{3n+1}\right)^{3n+1} -\ln(3n+1) \ \right]\ . $$