As I don't have at least 10 reputation so I cannot post formulates made by another web page like codecogs.
My question is that I need to calculate the geometric progression of (you can post this codeon codecogs and see what Im trying to ask)
$$ p_{k}=\frac{1}{3}\times 2^{-\begin{vmatrix} k \end{vmatrix}}$$
Where $k$ takes all integers numbers $\mathbb Z$.
The answer is
$$ \sum_{k\in \mathbb Z}^{\infty }p_k=p_{0}+2\sum_{k=1}^{\infty }p_{k}=\frac{1}{3}+2\cdot\frac{1}{3}\cdot \sum_{k=1}^{\infty }p_{k}=\frac{1}{3}+2\cdot\frac{1}{3}\cdot \frac{\frac{1}{2}}{1-\frac{1}{2}}=1 $$
Why do I have at the beginning and the answer
$$ p_{0}+2 $$
And the total answer it's not only
$$ \frac{1}{3}\frac{\frac{1}{2}}{1-\frac{1}{2}}$$
Thanks for your patience.
Notice how the sum is taken over all integers, not just $\mathbb Z_{\ge 0}$ as usual. Note also that $p_k = p_{-k}$ because of the definition. Therefore: $$ \begin{align} \sum_{k=-\infty}^\infty p_k &= p_0 + (p_{-1} + p_{-2} + \cdots) + (p_1 + p_2 + \cdots) \\ &= p_0 + (p_{1} + p_{2} + \cdots) + (p_1 + p_2 + \cdots) \\ \\ &= p_0 + 2\sum_{k=1}^\infty p_k \end{align} $$
Does it make sense?