Sum of geometric sequence How to get second system from the first one? I know that the sum of the geometric sequence is a(r^n-1)/(r-1), but I can not use this formula because of infinity and limitations on n.
  
 A: For example,
$$
\sum_{k=-\infty}^{-1}a^k = \sum_{k=1}^\infty a^{-k} = -1 + \sum_{k=0}^\infty \left(\frac 1 a\right)^k = -1 + \frac 1 {1-1/a} = \frac{1/a}{1-1/a}.
$$
A: Let's deal with $\sum_{k=-\infty}^{-n}a^k$, where $n\geq 0$. You can write $\sum_{k=-\infty}^{-n}a^k=\sum_{k=n}^{\infty}a^{-k}=\sum_{k=n}^{\infty}\left({1\over a}\right)^k$; you know that (for $0<r<1$, say)
$\sum_{k=0}^{n-1}r^k={1-r^n\over 1-r}$ and $\sum_{n=0}^{\infty}r^k={1\over 1-r}$.
Now,
$$
\sum_{k=0}^{\infty}\left({1\over a}\right)^k-\sum_{k=0}^{n-1}\left({1\over a}\right)^k={1\over 1-{1\over a}}-{1-\left({1\over a}\right)^n\over 1-{1\over a}}={\left({1\over a}\right)^n\over 1-{1\over a}},
$$ 
thus
$$
\sum_{k=-\infty}^{-n}a^k=\sum_{k=n}^{\infty}\left({1\over a}\right)^k={\left({1\over a}\right)^n\over 1-{1\over a}}={a^{-n}\over 1-{1\over a}}.
$$
If you prefer to write it with $n$ instead of $-n$, you have to keep in mind that $n<0$...
A: $$\sum _{k=-\infty }^n a^k=\sum _{k=-n}^{\infty } a^{-k}=\sum _{k=-n}^{-1} a^{-k}+\sum _{k=0}^{\infty } a^{-k}=\sum _{k=1}^n a^k+\sum _{k=0}^{\infty } a^{-k}=$$
$$ =\left(\frac{a^{n+1}-1}{a-1}-1\right)+\frac{1}{1-a}=\frac{a^{n+1}}{a-1}=\frac{a\cdot a^n}{a-1}=$$
divide numerator and denominator by $a$
$$=\frac{a^n}{1-1/a}$$
and the second one 
$$\sum _{k=-\infty }^{-1} a^k=\sum _{k=1}^{\infty } a^{-k}=\frac{1}{a-1}=\frac{1/a}{1-1/a}$$
