0

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.

Milten
  • 7,031
Limanido
  • 65
  • Welcome to MSE. If you can't post formulas yet, then just learn mathjax ;) Anyway, the third sum should have $2^{-k}$ instead of $p_k$. – Milten Nov 01 '19 at 20:05

1 Answers1

0

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?

Milten
  • 7,031
  • Thanks for your answer. So as pk=p-k, (p-1 + p-2 + p+3...) turn (p1 + p2 + p3..) and p0 = 1/3 its because k = 0 ? Another question is, why pk = p-k, where can I have an easy reading about it? Thanks. – Limanido Nov 01 '19 at 22:27
  • $p_0 = \frac13 2^{-|0|} = \frac13$. $p_{-k} = \frac13 2^{-|-k|} = \frac13 2^{-|k|} = p_k$. $p_{-k} = p_k$ only because of the definition of $p_k$ (it is an even function of $k$). – Milten Nov 01 '19 at 22:30