You are Here: Home >< Maths

# Quick question about convergence Watch

1. Here it is:

Does n(1-(1/n))^2 - n tends to 0 or to -2 ?
I find different reuslts each time

Thanks!
2. I get -2... I'm not sure how you'd get 0. Factorise it and see what cancels.
3. -2

how did you get 0?
4. well yeah I got -2 here (cause you can factorise and see it) but I think I confused myself when I had an arbitrary a>1 rather than 2, what I mean is:

n(1-(1/n))^a - n

Do you find that it tends to -a ? (and how)

thanks.
5. (Original post by hitheuk)
well yeah I got -2 here (cause you can factorise and see it) but I think I confused myself when I had an arbitrary a>1 rather than 2, what I mean is:

n(1-(1/n))^a - n

Do you find that it tends to -a ? (and how)

thanks.
Write and factorise it, then take the limit as (which is the same limit as )... what does this new limit look like?
6. (Original post by nuodai)
Write and factorise it, then take the limit as (which is the same limit as )... what does this new limit look like?
well theway I was doing is:

(1-(1/n))^a tends to 1 when n tends to infinity.
Then n( (1-(1/n))^a ) - n tends to n.1-n which tends to 0.

Where is the mistake ?
7. (Original post by hitheuk)
well theway I was doing is:

(1-(1/n))^a tends to 1 when n tends to infinity.
Then n( (1-(1/n))^a ) - n tends to n.1-n which tends to 0.

Where is the mistake ?
Your mistake is in taking the limit at two separate points. What you've done is take a limit (of the bracket), simplify (to give n-n=0) and then take another limit -- you can't do that. Alternatively, you might have taken a limit (of the bracket), then split the limits to give , and tried to subtract infinity from itself... which is never a good idea.

When you're taking limits, you can only use the rules and when and both converge. Here, this doesn't happen, since doesn't converge as .
8. (Original post by nuodai)
Your mistake is in taking the limit at two separate points. What you've done is take a limit (of the bracket), simplify (to give n-n=0) and then take another limit -- you can't do that. Alternatively, you might have taken a limit (of the bracket), then split the limits to give , and tried to subtract infinity from itself... which is never a good idea.

When you're taking limits, you can only use the rules and when and both converge. Here, this doesn't happen, since doesn't converge as .
Ok thanks.

(Original post by nuodai)
Write and factorise it, then take the limit as (which is the same limit as )... what does this new limit look like?

So I get (1/h) [1-(1+h)^a] and then the first bit goes to infinity and the second bit to 0 when h tends to 0 ...
9. (Original post by hitheuk)
Ok thanks.

So I get (1/h) [1-(1+h)^a] and then the first bit goes to infinity and the second bit to 0 when h tends to 0 ...
Would it look more familiar if you wrote it as:
?
10. (Original post by nuodai)
Would it look more familiar if you wrote it as:
?
Derivative of the function - (1+x)^a at zero ! So the answer is - a
11. (Original post by hitheuk)
Derivative of the function - (1+x)^a at zero ! So the answer is - a
Correct!

Alternatively you could have used a Taylor expansion for , but this way's more fun.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 29, 2010
Today on TSR

### Degrees to get rich!

... and the ones that won't

### Women equal with Men?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.