Approximate the following series $\sum_{k=11}^{\infty}\frac{1}{k\ 3^k}$ I am trying to estimate the following serie.
$$\sum_{k=11}^{\infty}\frac{1}{k\ 3^k}$$
I am thinking about comparison, but I am stuck.
It doesn't give a clear answer. It does look like a combined arithmetic and geometric serie.

We know that $$\ln (1-x) = - \sum_{k=1}^\infty \frac{x^k}{k}$$
By the Taylor series expansion. Hence $$-\ln (1-x) =\sum_{k=1}^\infty \frac{x^k}{k} $$
Now we set $x = 1/3$ :
$$-\ln (1-1/3) =\sum_{k=1}^\infty \frac{(1/3)^k}{k} = \sum_{k=1}^\infty \frac{1}{3^k\cdot k}$$
Therefore the sum from $k=1$ to $\infty$ = $-\ln (2/3) = \ln (3/2)$. However your sum starts at $k=11$ so we need to remove the first $10$ terms of the sum :
$$
\sum_{k=11}^\infty \frac{1}{3^k\cdot k} = \ln (3/2) - \sum_{k=1}^{10}\frac{1}{3^k\cdot k}
$$
This can be evaluated using WA.
