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.
 A: Hint: Write the Taylor series of $$\log \Big(1 - \frac{1}{3}\Big)$$
ant subtract by $s_{10}$.
A: 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.
A: Considering the series
$$\sum_{k=0}^\infty x^{k+10}=x^{10}(1-x)^{-1}$$
we obtain by integration
$$\sum_{k=0}^\infty\frac{\left(\frac13\right)^{k+11}}{k+11}=\int_{x=0}^{1/3}x^{10}(1-x)^{-1}dx=B\left(\frac13;11,0\right)$$
where $B$ is the incomplete Beta function.
(As shown in other posts, a closed-form with elementary functions exists.)
A: The question doesn't specify what is meant by "approximate", but I would note that none of the current answers lead directly to an approximation—yes, we can directly calculate the sum to be equal to $\ln(\frac{3}{2}) - \frac{20111503}{49601160}$ but how exactly are we planning to estimate that?
This is a job for Taylor's theorem.  The sum in question is just $R_{10}(\frac{1}{3})$, where $R(x) = f(x) - P_{10}(x)$ is the tenth remainder in the Taylor series for $f(x) = -\ln(1-x)$.
Taylor's theorem (specifically, the Lagrange form of the remainder) tells us that $R_{10}(\frac{1}{3}) = \frac{f^{(11)}(\zeta)}{11!}\left( \frac{1}{3} \right)^{11}$ for some $\zeta\in [0,\frac{1}{3}]$.  Since $f^{(11)}(x) = \frac{10!}{(1-x)^{11}}$, with maximum value $10! \cdot \left(\frac{3}{2}\right)^{11}$ in this interval, we arrive at:
$$R_{10}(\frac{1}{3}) \leq \frac{1}{11} \cdot \left(\frac{1}{2}\right)^{11} < 5\cdot 10^{-5}$$
The true value is about $7\cdot 10^{-7}$, so this is correct to four decimal places.
