Here is the homework question I am working on:
Evaluate (as a real number) the series $$L(1, \chi_3) = \sum_{n=1}^\infty \frac{\chi_3(n)}{n},$$ where $\chi_3$ is the unique nontrivial Dirichlet character mod $3$.
My work is as follows:
We have that \begin{align*} L(1, \chi_3) = \sum_{n=1}^\infty \frac{\chi_3(n)}{n} &= \frac{\chi_3(1)}{1} + \frac{\chi_3(2)}{2} + \frac{\chi_3(4)}{4} + \frac{\chi_3(5)}{5} + \frac{\chi_3(7)}{7} + \cdots \\ &= 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{5} + \frac{1}{7} - \frac{1}{8} + \frac{1}{10} - \frac{1}{11} + \cdots \\ &= \left(1 + \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{8} + \frac{1}{11} + \cdots\right) \\ %&= 1 + \sum_{n=1}^\infty\frac{1}{3n + 1} - \sum_{n=1}^\infty \frac{1}{3n - 1} \\ &= 1 + \sum_{n=1}^\infty\frac{1}{3n + 1} - \frac{1}{3n - 1}. \end{align*} Now, we define $$f(x) = 1 + \sum_{n=0}^\infty\frac{x^{3n + 1}}{3n + 1} - \frac{x^{3n - 1}}{3n - 1}$$ and differentiate termwise to find that \begin{align*} f'(x) &= \sum_{n=0}^\infty x^{3n} - x^{3n - 2} \\ &= \frac{x^3}{1 - x^3} - \frac{x}{1 - x^3} \\ &= \frac{x(x^2 - 1)}{1 - x^3} \end{align*} so \begin{align*} f(1) &= 1 + \int_0^1 \frac{x(x^2 - 1)}{1 - x^3} \\ &= 1 + \left(\frac{\sqrt3 \pi}{9} - 1\right) \\ &= \frac{\sqrt3 \pi}{9} \end{align*}
Is this the correct answer and method?
Your answer and method are indeed correct.
Here is another simple method which gives the same value. First, note that
$$\chi_3(n)=-\frac{i\sqrt{3}}3(e^{i2n\pi/3}-e^{-i2n\pi/3})$$
We therefore have that
\begin{equation} f(x):=\sum_{n=1}^\infty \frac{\chi_3(n)}{n}x^n=-\frac{i\sqrt{3}}3\left[\sum_{n=1}^\infty \frac{e^{i2n\pi/3}}{n}x^n-\sum_{n=1}^\infty \frac{e^{-i2n\pi/3}}{n}x^n\right]=\frac{i\sqrt{3}}3[\log(1-e^{i2\pi/3}x)-\log(1-e^{-i2\pi/3}x)] \end{equation}
for $|x|<1$. Since $\sum_{n=1}^\infty \frac{\chi_3(n)}{n}$ converges via the alternating series test, then by Abel's theorem,
\begin{equation} \sum_{n=1}^\infty \frac{\chi_3(n)}{n}=\lim_{x\rightarrow 1^-}f(x)=\frac{i\sqrt{3}}3[\log(1-e^{i2\pi/3})-\log(1-e^{-i2\pi/3})]=\frac{i\sqrt{3}}3\log(e^{-i2\pi/6})=\frac{\sqrt{3}\pi}{9} \end{equation} as desired.