never mind, just sparked in my head that this is a very simple induction problem for cases where n_0 is greater than or equal to 2. =)
i am stuck on proving this property of a series:
let p_n=sigma (from k=1 to n) (1/k) and p=sigma (from k=1 to infinite) (1/k).
Prove for all N, a natural number, there exists a n_0 in the naturals such that abs (p_n-p_n-1)<=1/N for all n>=n_0 yet abs (p-p_n)> 1/N for all n>=n_0.
essentially one would need to show that abs (1/n) <= 1/N and that sigma(from k=n+1 to infinite) (1/n)>=1/N. how would we choose our n_0 in a way that this would work? thanks!
Let:
and suppose:
exists.
Then for any integer :
so for any :
Now consider:
The first series comprises terms all greater than or equal to so is greater than or equal to , and the second series is similarly greater than or equal to and the third is greater thab zero, so:
So putting will do.
However note that this is all predicated on the existance of , but there is no such as the harmonic series diverges.
RonL
No I mean 1/(2n), since this is the samllest term in that finite series, and by construstion there are n terms all greater than or equal to the smallest.
No I am using the series given, it just happens to be the harmonic series
This is an adaptation of one method of proving that the harmonic series is divergent.
RonL