# Sequence of real numbers | Proof of convergence

• January 23rd 2010, 08:45 PM
kingwinner
Sequence of real numbers | Proof of convergence

Assuming the truth of part a, I proved part b.
But now I am struggling with proving parts a & c.
Part a seems true intuitively. The sqaure root of a number between 0 and 1 is will be larger than that number, and if we take more and more square roots, it will get close to 1, and then if we add two numbers that are close to 1, it must be ≥1.
But how can we write a FORMAL proof of it? How can we construct N and demonstrate exactly that there exists an N such that n≥N => a_n ≥1?

Can someone help me, please?
Any help is much appreciated!

[note: also under discussion in Math Links forum]
• January 24th 2010, 04:24 AM
Prometheus
for part a, if $0 then as you said $\sqrt{a_n}>a_n$, but then by definition $a_{n+1}>\sqrt{a_n}$ and you have an increasing monotone series. if it was bounded by one, it would converge. and then you can take 2 elements close enough the the convergence point and show that the next point is above it.

if the series converge to x then you'll have $x = 2 \sqrt{x} \rightarrow x(4-x)=0$, so x=4 or x=0.

you showed in part b something about the distance of the elements of the series from 4. if you can show this approaches to zero as n goes to infinity, then the series converges to 4.
• January 24th 2010, 01:11 PM
kingwinner
I'm sorry, I don't exactly follow your argument. Thanks for your help though.

But I think if a(n) and a(n+1) are positive then we can surely pick an m such that 1/(2^m) is less than them both: just select a sufficiently large m, e.g. select m such that 2^m >= 1/min{a(n),a(n+1)}.

But how can I find N such that n>=N implies a(n)>=1 ?

Any help is appreciated!
• January 25th 2010, 02:22 PM
kingwinner
OK, using a different argument, I finishing proving part a, and using part a, I proved part b.

Now I'm stuck on part c. May someone help me with part c, please?