Limit of a sequence

1. Jun 26, 2003

KLscilevothma

It isn't a homework problem but I think I better post it here instead of Mathematics forum, since it belongs to "exam help".

Prove that for any positive real numbers a and b,
lim [(an+b)1/n-1] = 0
n->inf

I don't need to use things like |a-b|<epsilon. A simple way will do. I know it's an easy question but I don't know where to start. Could someone please help.

2. Jun 26, 2003

Tom Mattson

Staff Emeritus
This one just screams "L'Hopital!"

First, rearrange it to:

lim(an+b)1/n=1
n-->&infin;

Then take the natural log of both sides to get:

lim ln(an+b)/n=0
n-->&infin;

This goes to &infin;/&infin;, which is an indeterminate form and ripe for L'Hopital's rule.

3. Jun 26, 2003

KLscilevothma

LOL, thanks Tom and L'hopital

lim ln(an+b)/n
n->[oo]

= lim a/(an+b)
n->[oo]
=0

4. Jun 27, 2003

KLscilevothma

Oh sorry, I forgot to mention
(an+b)1/n-1
is a sequence, not a function. I think L'hopital's rule applies to differentiable functions only.

Perhaps I better rephase the question a bit.
A sequence {an} is defined by (an+b)1/n-1
Prove that
lim (an+b)1/n-1 = 0
n->inf
(a and b are real numbers and n is a positive integer)

5. Jun 27, 2003

HallsofIvy

Staff Emeritus
It is true that L'hopital's rule applies to functions rather than sequences.

However, IF we can convert a sequence an to a function f(x) (we can't if the sequence involves things like n! or "floor" or "ceiling" that can't be written simply as a continuous function), then f(x)-> L, an-> L. The other way doesn't necessarily work- the function might not have a limit, depending on how it is defined for non-integer values.

6. Jun 27, 2003

KLscilevothma

So we can treat a sequence as a function if it is an "elementary" one like the one I posted, and can apply L'hopital's rule, is it correct?