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Quick question about convergence

Here it is:

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

Thanks!
Reply 1
Avatar for nuodai
nuodai
17
I get -2... I'm not sure how you'd get 0. Factorise it and see what cancels.
-2

how did you get 0?
Reply 3
Avatar for hitheuk
hitheuk
OP
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.
Reply 4
Avatar for nuodai
nuodai
17
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 n=−1hn=-\dfrac{1}{h} and factorise it, then take the limit as h→0h \to 0 (which is the same limit as n→∞n \to \infty)... what does this new limit look like?
Reply 5
Avatar for hitheuk
hitheuk
OP
Original post by nuodai
Write n=−1hn=-\dfrac{1}{h} and factorise it, then take the limit as h→0h \to 0 (which is the same limit as n→∞n \to \infty)... 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 ?
Reply 6
Avatar for nuodai
nuodai
17
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 lim⁡n−lim⁡n\lim n - \lim n, and tried to subtract infinity from itself... which is never a good idea.

When you're taking limits, you can only use the rules lim⁡(anbn)=lim⁡anlim⁡bn\lim (a_nb_n) = \lim a_n \lim b_n and lim⁡(an+bn)=lim⁡an+lim⁡bn\lim (a_n + b_n) = \lim a_n + \lim b_n when ana_n and bnb_n both converge. Here, this doesn't happen, since an=na_n=n doesn't converge as n→∞n \to \infty.
Reply 7
Avatar for hitheuk
hitheuk
OP
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 lim⁡n−lim⁡n\lim n - \lim n, and tried to subtract infinity from itself... which is never a good idea.

When you're taking limits, you can only use the rules lim⁡(anbn)=lim⁡anlim⁡bn\lim (a_nb_n) = \lim a_n \lim b_n and lim⁡(an+bn)=lim⁡an+lim⁡bn\lim (a_n + b_n) = \lim a_n + \lim b_n when ana_n and bnb_n both converge. Here, this doesn't happen, since an=na_n=n doesn't converge as n→∞n \to \infty.


Ok thanks.

Original post by nuodai
Write n=−1hn=-\dfrac{1}{h} and factorise it, then take the limit as h→0h \to 0 (which is the same limit as n→∞n \to \infty)... 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 ...
Reply 8
Avatar for nuodai
nuodai
17
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:
−lim⁡h→0(1+h)a−1ah\displaystyle -\lim_{h \to 0} \dfrac{(1+h)^a - 1^a}{h} ?
Reply 9
Avatar for hitheuk
hitheuk
OP
Original post by nuodai
Would it look more familiar if you wrote it as:
−lim⁡h→0(1+h)a−1ah\displaystyle -\lim_{h \to 0} \dfrac{(1+h)^a - 1^a}{h} ?


Derivative of the function - (1+x)^a at zero ! So the answer is - a
Reply 10
Avatar for nuodai
nuodai
17
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 (1−1n)a(1-\frac{1}{n})^a, but this way's more fun.

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