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

[penguin007]

Homework Statement

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?

Homework 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.

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.

 

[Mark44]

Homework Statement

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?

Homework 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.

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.

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/n^a in the neighborhood of +infinity"?

 

[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.