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 $\frac{9}{4}.$

Firstly, what you have is a limit of a finite sum, and the limit is a series:
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{k+1}{3^k}:=\sum_{k=1}^\infty \frac{k+1}{3^k}$$
Now, to the question you asked: there are a few tricks you can use:
1) You know that $\sum_{k=1}^nx^k=\frac{1-x^{n+1}}{1-x}-1$ (geometric progression).
2) That implies, by deriving both sides, that $\sum_{k=1}^nkx^{k-1}=\frac{-(n+1)x^{n}(1-x)+(1-x^{n+1})}{(1-x)^2}=\frac{nx^{n+1}-(n+1)x^{n}+1}{(1-x)^2}$
3) From here you get that $\sum_{k=1}^n(k+1)x^k=\frac{1-x^{n+1}}{1-x}-1+x\frac{nx^{n+1}-(n+1)x^{n}+1}{(1-x)^2}=\frac{(n+1)x^{n+2}-(n+2)x^{n+1}-x^2+2x}{(1-x)^2}$.
4) Now plug in $x=\frac{1}{3}$. As $n\to\infty$, you have $x^n\to 0$. This implies that $$\lim_{n\to\infty} \sum_{k=1}^n \frac{k+1}{3^k}=\frac{-\left(\frac{1}{3}\right)^2+2\cdot\frac{1}{3}}{\left(\frac{2}{3}\right)^2}=\frac{5}{4}$$
