<s>[INST] 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 $\frac{9}{4}.$
 [/INST] For summing $\sum\frac{k}{3^k}$ use the following formula: $$(1-x)^{-2} = 1 + 2x + 3x^{2} + 4x^{3} + \cdots \qquad \Bigl[\because \small (1-x)^{-n} = 1+nx +\frac{n\cdot (n-1)}{2!}\cdot x^{2} + \cdots \Bigr]$$ Multiplying the above equation by $x$ and then putting $x=\frac{1}{3}$ we have $$\frac{1}{3} + \frac{2}{9}+\frac{3}{27} + \frac{4}{3^{4}} + \cdots  = \frac{1}{3}\Bigl(1-\frac{1}{3}\Bigr)^{-2} = \frac{3}{4} \qquad\quad \cdots (1)$$
Also you know that $$\sum\limits_{k=1}^{\infty} \frac{1}{3^k}= \frac{\frac{1}{3}}{1-\frac{1}{3}} =\frac{1}{2} \qquad\qquad \cdots (2)$$
Add equations $(1)$ and $(2)$ to get your answer.
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