# Equivalent of 1/n^(a) in the neighborhood of +infinity

1. Oct 10, 2009

### penguin007

1. The problem statement, all variables and given/known data
How can one find an equivalent of 1/n^(a) in the neighborhood of +infinity?
And what's the method in general to get an asymptotical development of a sequence?

2. Relevant equations
The result is:
1/n equivalent to ln(n+1)-ln(n)
1/n^a equivalent to (1/a-1)*(1/n^(a-1)-1/(n+1)^(a-1) if a is different from 1.

3. The attempt at a solution
I tried to use Taylor Young with f:=x->x in the neighborhood of 0 (this is useless)

Thanks a lot for your help.

2. Oct 10, 2009

### Staff: Mentor

No one has responded to your question all day, which might mean that they are as confused about what you are asking as I am. What do you mean "an equivalent of 1/na in the neighborhood of +infinity"?

3. Oct 11, 2009

### penguin007

I mean an asymptotical expansion: for instance ln(n+1)-ln(n)=ln(1+(1/n))=(1/n)+o(1/n) (because 1/n is about 0 when n tends to infinity). Therefore 1/n is equivalent to ln(n+1)-ln(n) in the neighborhood of infinity.
My question is how do we get this equivalent?

Thanks.