Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

It is simple for Mathematica to find an asymptotic expansion for $\frac{1}{-1+p}$ as $p \rightarrow \infty$. However, if we want to restrict $p$ to be an integer and also include some terms that simplify for integer p (but not generally) then Series by itself will no longer do the job (without some prodding, I expect).

Is there a simple way to use Series to find the asymptotic expansion of $$\frac{1}{(-1)^p+p}$$ as $p \rightarrow \infty$, assuming integer $p$?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Simplicity is arguably in the eye of the beholder, but the following code gets you closer to $\sum_{k=0}^{n}(-1)^k p^{-(k+1)}$

ps = Series[1/((-1)^p + p), {p, \[Infinity], 1}];
psEven = Simplify[ps,  Assumptions -> p \[Element] Integers \[And] Mod[p, 2] == 0]
psOdd = Simplify[ps,  Assumptions -> p \[Element] Integers \[And] Mod[p, 2] == 1]

for either even p or odd p.

share|improve this answer
1  
Using Simplify as a post-processor to Series seems like a good way to go about this. –  Daniel Lichtblau Jan 31 at 16:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.