sequences limits and cauchy sequences

ibc

1. The problem statement, all variables and given/known data
prove or refute:

if lim(a(2n)-a(n)=o , then a(n) is a cauchy sequence


2. Relevant equations



3. The attempt at a solution
I need to prove that for every m,n big enough a(m)-a(n)<epsilon
so I know for all m and n I can say m=l*n, lim(a(m)-a(n))=lim(a(n*l)-a(n*l/2) +a(n*l/2) -a(n*l/4)..........+a(2n)-a(n)), which is the sum of alot of zeros, though if I take m to be 2^n or something like that, I get an inifinite amount of zeros, so I don't know what I can do with that.
so I tried to find a sequence which contredicts it, though couldn't find any




1. The problem statement, all variables and given/known data
prove or refute:
if |a(n+1)-a(n)|<9/10*|a(n)-a(n-1)|
then a(n) is a cauchy sequence


2. Relevant equations




3. The attempt at a solution
well fromt he equation I can get:
(|a(n+1)-a(n)|)/(|a(n)-a(n-1))<9/10<1
so what it gives me is that all that in the power of n is going to zero, which means is a cauchy sequence, thoguh I don't see how it helps me =\

Dick

Yes, you want a counterexample for the first one. It's going to have to be a function that grows REALLY slowly. What kind of functions can you think of that do that? For the second one for n<m, a(n)-a(m)=(a(n)-a(n+1))+(a(n+1)-a(n+2))+(a(n+2)-a(n+3))+ ... (a(m-1)-a(m)). Think you can maybe bound that sum by a geometric series?

ibc

I know I need something that grows really really slow, but I couldn't find any
and we can't use functions yet in the course, so it'd have to be a sequence/series
the slowest one I could think of is 1/1+1/2+1/3+1/4+...... which is good to refute the another question, which was "if lim(a(n+1)-a(n)=o , then a(n) is a cauchy sequence", though for this one it's not helpful.

Dick

a(n)=log(n) grows pretty slowly, but not quite slowly enough. Try sqrt(log(n)).