# Show sequence is convergent

• October 30th 2013, 11:33 AM
sakuraxkisu
Show sequence is convergent
Hi, my questions are:

1)Circle the convergent sequences in the following list (note: I haven't included the whole list), and find the limit in case the sequence converges.
1a) n^(1/n)
b) 0.9, 0.99, 0.999, ...

For 1a) : By calculating the first few terms, I found that as n tends to infinity, n^(1/n) tends to 1, so I know that limit is 1, but I need to show this formally using the definition of convergence: for all epsilon, e>0 there exists a natural number N such that for all n>N, |an - L|<e.

All I've done so far is say that |n^(1/n) - 1| , however I'm not sure which N to choose so that for all n>N, this absolute value will be less than epsilon e.

For 1b) I know that the sequence tends to 1 as n tends to infinity, however I'm not sure how to prove this formally. I thought of using the theorem that any bounded and monotonically increasing or decreasing sequence is convergent, and I know that I'd need to use proof by induction to show both, but again I'm not quite sure how to do this.

Any help would be greatly appreciated! :)
• October 30th 2013, 04:02 PM
chiro
Re: Show sequence is convergent
Hey sakuraxkisu.

For 1b) the series is basically 1 - (0.1)^n which is monotonically decreasing. For 1a) you might want to consider another sequence that is bigger than n^(1/n) but still convergent and easier to prove and show that since n^(1/n) < other-sequence then n^(1/n) converges as well.
• October 30th 2013, 04:04 PM
HallsofIvy
Re: Show sequence is convergent
If you have, and can use, the theorem "If a sequence is bounded and monotone, then it converges", for the second sequence it is sufficient to note that 1- .1^n is increasing and always less than 1. It is obviously less than 1 because .1<1 so .1^n< 1 for all n. It is increasing because .1< 1 so, multiplying both sides by .1^n, .1^{n+1}< .1^n. Then multiplying both sides by -1, -.1^{n+1}> -.1^n so 1- .1^{n+1}> 1- .1^n.

As for $a_n= n^{1/n}$, I would suggest looking at $ln(a_n)= (1/n)ln(n)$. Can you show that ln(n)/n goes to 0?
• October 31st 2013, 07:13 AM
sakuraxkisu
Re: Show sequence is convergent
Thank you both for your help!

For 1a, here's what I've done, following from the advice you both gave:
Let tn=ln(an)=ln(n)/n.
To show ln(n)/n is convergent with limit 0, let epsilon e>0 be given and choose N>1/(e^2). Then, for all n>N, and using the fact that ln(n)< sqrt(n) for all natural numbers n,
|ln(n)/n - 0|=|ln(n)/n|=ln(n)/n < sqrt(n)/n= 1/sqrt(n) < 1/sqrt(N) < e

Is this fine? I'm unsure as to whether I can use the fact that ln(n) < sqrt(n), as it hasn't been mentioned explicitly in lectures and is something I found out online.
• October 31st 2013, 07:46 AM
sakuraxkisu
Re: Show sequence is convergent
Also, for 1b), if I wanted to use the definition of convergence to prove the sequence 1-(0.1)^n converges, would this be alright?
an=1-(0.1)^n. Let epsilon e>0 be given, and choose N>log to the base 10 of 1/e , such that for all n>N,
|1-(0.1)^n -1|=(0.1)^n=(1/10)^n=1/(10^n) < 1/(10^N) < e