What function does $\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$ represent, evaluated at some number $x$? I need to know what the function $$\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$$ represents evaluated at a particular point.
For example if the series given was $$\sum \limits_{n=0}^{\infty}\frac{3^n}{n!}$$ the answer would be $e^x$ evaluated at $3$.
Yes, this is homework, but I'm not looking for any handouts, any help would be greatly appreciated.

Take $f(x) = \displaystyle \sum_{n=1}^\infty \frac{x^n}{n}$.
Then,
$f^\prime (x) = \displaystyle \sum_{n=1}^\infty \frac{n x^{n-1}}{n} = \sum_{n=0}^\infty x^n$.
The last expression is a geometric series and, as long as $x < 1$, it can be expressed as
$f^\prime (x) = \displaystyle \frac{1}{1-x}$.
Therefore,
$f(x) = - \ln | 1 - x | + \kappa$
Where $\kappa$ is a constant. But if you take the original expression for $f(x)$, you can see that $f(0) = 0$ and, therefore, $\kappa = 0$.
So $f(x) = -\ln | 1 - x |$.
The answer to your question is just $f \left(\frac{1}{3} \right)$.
You can also obtain this result by Taylor expanding $\ln ( 1 - x )$.
