Thread: 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!

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.

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 $\displaystyle a_n= n^{1/n}$, I would suggest looking at $\displaystyle ln(a_n)= (1/n)ln(n)$. Can you show that ln(n)/n goes to 0?

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.

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