<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] Expanding your problem:
$$ \sum\limits_{k = 0}^\infty  {\frac{{k + 1}}{{3^k }}} = \frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \frac{4}{3^3} + \dots $$
$$ = 1 + \left (\frac{1}{3} + \frac{1}{3} \right) +  \left(\frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{3^2} \right) +  \left(\frac{1}{3^3} + \frac{1}{3^3}+ \frac{1}{3^3}+ \frac{1}{3^3}\right) + \dots$$
This can be grouped into:
$$ = \left(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$
$$ \left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$
$$ \left(\frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$
$$ \left(\frac{1}{3^3} + \dots\right) + \dots $$
Using the fact that $ S = \sum_{n=0}^{\infty} \frac{1}{3^n} = \frac{3}{2}$:
$$ = \frac{3}{2} + $$
$$ \frac{3}{2} - (1) + $$
$$ \frac{3}{2} - \left(1 + \frac{1}{3} \right) + $$
$$ \frac{3}{2} - \left( 1 + \frac{1}{3} + \frac{1}{3^2} \right ) + \dots $$
The partial sum $S_k$ is computed as: $S_k = \sum_{n=0}^k \frac{1}{3^n} = \frac{3}{2} - \frac{1}{2}\left(\frac{1}{3}\right)^k$
Hence,
 $$ = \frac{3}{2} + \left(\frac{3}{2} - S_0 \right) + \left(\frac{3}{2} - S_1 \right) + \left(\frac{3}{2} - S_2 \right) \dots$$
$$ = \frac{3}{2} + \frac{1}{2} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \dots \right) $$
$$ = \frac{3}{2} + \frac{1}{2}S = \frac{3}{2} + \frac{1}{2} \frac{3}{2}$$
$$ = \mathbf{\frac{9}{4}}$$
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