Limit of this series: $\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$? Given a series, how does one calculate that limit below? I noticed the numerator is an arithmetic progression and the denominator is a geometric progression — if that's of any relevance —, but I still don't know how to solve it.
$$\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$$
I did it "by hand" and the result should be $\frac{9}{4}.$

divide that formular like this.
$$\sum_{k=1}^{\infty}\left (\frac{k}{3^k}+\frac{1}{3^k}\right )$$
then $$\sum_{k=1}^{\infty}\frac{k}{3^k}+\frac{\frac{1}{3}}{1-\frac{1}{3}}=\sum_{k=1}^{\infty}\frac{k}{3^k}+\frac{1}{2}$$
power series
$$ \sum_{k=1}^{\infty}\frac{k}{3^k}$$
let $$ \sum_{k=1}^{\infty}\frac{k}{3^k}=S$$
Then,
$$ S=1\times \frac{1}{3}+2\times \frac{1}{3^2}+3\times \frac{1}{3^3}+\cdots \cdots $$
$$\frac{1}{3}S=1\times \frac{1}{3^2}+2\times \frac{1}{3^3}+3\times \frac{1}{3^4}+\cdots \cdots $$
$$S-\frac{1}{3}S=1\times \frac{1}{3}+(2-1)\times \frac{1}{3}+(3-2)\times \frac{1}{3}\cdots \cdots =\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}$$
Answer is
$$\therefore \frac{2}{3}S=\frac{1}{2},S=\frac{3}{4}$$
$$\frac{3}{4}+\frac{1}{2}=\frac{5}{4}$$
