Math Help - stuck on proving properties of a series

1. stuck on proving properties of a series

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!

2. 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. =)

3. i was trying to show that this inequality holds for n+1 case, but found out that theres no way to bound this by the n case which we can assume true by induction hypothesis. Help? thanks. ((

4. Originally Posted by squarerootof2
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!
Hmm well as far as I can see $n_0 = N$, works wonderfully well

5. i can see that we are going to need proof by induction, but can you find a way to prove the second condition for the n+1 case? i cant seem to find a way that works. - thanks

6. Originally Posted by squarerootof2
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).
Am I missing something here?

You define:

$p=\sum_{k=1}^{\infty} \frac{1}{k}$

That is $p$ is the sum of the harmonic series, but the harmonic series is divergent!

Ahh... now I see what is going on, you are proving that the harmonic series is divergent by contradiction... or meybe not

RonL

7. Originally Posted by squarerootof2
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:

$p_n=\sum_{k=1}^n \frac{1}{k}$

and suppose:

$p_n=\sum_{k=1}^{\infty} \frac{1}{k}$

exists.

Then for any integer $n$:

$p_n-p_{n+1}=\frac{1}{n}$

so for any $n\ge N$:

$p_n-p_{n+1}=\frac{1}{n} \le \frac{1}{N}.$

Now consider:

$
p-p_n=\sum_{k=n+1}^{\infty}\frac{1}{k}=\sum_{k=n+1}^ {2N}\frac{1}{k}+\sum_{k=2n+1}^{4n}\frac{1}{k}+\sum _{k=4n+1}^{\infty}\frac{1}{k}
$

The first series comprises $n$ terms all greater than or equal to $1/(2n)$ so is greater than or equal to $1/2$, and the second series is similarly greater than or equal to $1/2$ and the third is greater thab zero, so:

$
p-p_n=\sum_{k=n+1}^{\infty}\frac{1}{k} \ge 1 \ge \frac{1}{N}
$

So putting $n_0=N$ will do.

However note that this is all predicated on the existance of $p$, but there is no such $p$ as the harmonic series diverges.

RonL

8. one question, do u mean (1/2)n instead of 1/(2n) when u justify the first series being greater than 1/2? and also, this is from the definition of harmonic series, right?

9. Originally Posted by squarerootof2
one question, do u mean (1/2)n instead of 1/(2n) when u justify the first series being greater than 1/2? and also, this is from the definition of harmonic series, right?
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