Lol or not Lol? Here is the log

Maclurin series of $\ln (1-x) = - \displaystyle \sum_{n=1}^\infty \frac{x^n}{n}$ .

Substituting $x = \dfrac{1}{3}$ , we get $\displaystyle \sum_{n=1}^\infty \frac{1}{3^n n} = - \ln \left(1-\frac{1}{3}\right) = - \ln \left(\frac{2}{3}\right) = \ln \left(\frac{3}{2}\right)$

$\implies a + b = \boxed{5}$

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$\underline{\text{Another Method}}$

$\begin{aligned} \sum_{n=1}^\infty \frac{\color{#3D99F6}{1}}{\color{#3D99F6}{3}^n n} & = \sum_{n=1}^\infty \frac{\color{#3D99F6}{x}^n}{n} \quad \quad \small \color{#3D99F6}{\text{Let } x = \frac{1}{3}} \\ & = \sum_{n=1}^\infty \int x^{n-1} dx = \int \sum_{n=1}^\infty x^{n-1} dx = \int \sum_{n=0}^\infty x^{n} dx \\ & = \int \sum_{n=0}^\infty x^{n} dx = \int \frac{1}{1-x} dx \\ & = - \ln (1-x) \quad \quad \small \color{#3D99F6}{\text{Putting back } x = \frac{1}{3}} \\ & = \ln \left( \frac{3}{2}\right) \end{aligned}$

$\implies a + b = \boxed{5}$

My solution is similar to Compañero Guillermo's.

We have the Taylor Series $\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$ for $|x|<1$ . Evaluating this at $x=\frac{1}{3}$ gives $S=-\ln\left(\frac{2}{3}\right)=\ln\left(\frac{3}{2}\right)$ . The answer is $\boxed{5}$

$\displaystyle \ln (1 + x) = \sum_{n = 1}^\infty \frac{(-1)^{n + 1} x^n}{n}, \text{ if } |x| < 1$ ,for researching this formula see next pragraph. We know $\displaystyle \sum_{n = 0}^\infty x^n = \frac{1}{1 - x}, \text{ if } |x| < 1$ Integrating,and using Abel theorem or Weierstrass theorem for uniform convergence of continuous functions we get $\displaystyle \sum_{n = 0}^\infty \frac{x^{n + 1}}{n + 1} = - \ln (1 - x) + C, \text{ with C a constant and } |x| < 1$ Evaluating at $x = 0$ we get $C = 0$ . This implies $\displaystyle \sum_{n = 1}^\infty \frac{1}{3^n n} = - \ln (\frac{2}{3}) = \ln (\frac{3}{2}) \Rightarrow a + b = 5$