I would like to find an expression where one adds to $\gamma$ smaller and smaller fractions of $1/\gamma$, like:
$p=\gamma+1/\gamma+1/\gamma^3+1/\gamma^5+...+/\gamma^n$
where n is always odd. I tried
$\gamma+\sum_{i=1+2n}^n\gamma^{-n}$
But in Mathematica online it does not give the right number. What is the correct way here?
Thank you
$p-\gamma =1/ \gamma + 1/ \gamma^{3} +...+1 / \gamma^n $. $\\$ The RHS is a geometric finite sum where ratio is $\gamma^{-2}$.
It is mostly a geometric series in $q=1/\gamma^2$. $$ \sum_{k=0}^N q^k = \frac{1-q^{N+1}}{1-q} \to \frac{1}{1-q} $$ for $\lvert q \rvert < 1$ and $N \to \infty$.
Then if $\gamma \ne 0$ and $N \ge 0$: \begin{align} p = \gamma + \frac{1}{\gamma} + \frac{1}{\gamma^3} + \cdots + \frac{1}{\gamma^{2N-1}} &= \gamma \sum_{k=0}^N \left(\frac{1}{\gamma^2}\right)^k \\ &= \gamma \frac{1-(1/\gamma^2)^{N+1}}{1-(1/\gamma)^2} \\ &= \gamma \frac{1-1/\gamma^{2N+2}}{1-1/\gamma^2} \\ &= \gamma \frac{1-1/\gamma^{n+3}}{1-1/\gamma^2} \\ &\to \gamma \frac{1}{1-1/\gamma^2} \end{align} for $\lvert q \rvert = 1/\gamma^2 < 1 \iff \lvert\gamma\rvert > 1$ and $N \to \infty$. We had $n = 2N-1$.
