Geometric Series
Warning
This post is more than a year old. Information may be outdated.
In Probability,
$$ \sum\limits_{j=k}^{N} a^j = {{a^k - a^{N+1} \over {1-a}}} $$
$$
\text{if} |a| < 1, ~\sum\limits_{j=k}^{\infty} a^j = \lim_{N \to \infty} \sum\limits_{j=k}^{N} a^j = {{a^k \over {1-a}}}
$$
$$ \sum\limits_{j=1}^{\infty} a^j = {a \over {1-a}} $$
$$ \sum\limits_{j=0}^{\infty} a^j = {a \over {1-a}} $$
$$ \sum\limits_{j=1}^{\infty} ja^j= {a \over {(1-a)^2}} $$
$$ \sum\limits_{j=0}^{\infty} ja^j= {a \over {(1-a)^2}} $$
$$ \sum\limits_{j=1}^{\infty} j^2 a^j= {{a + a^2} \over {(1-a)^2}} $$
$$ \sum\limits_{j=0}^{\infty} j^2 a^j= {{a + a^2} \over {(1-a)^2}} $$