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Divergent Harmonic Series, Convergent P-Series (Cauchy sequences)

  • Thread starter PingPong
  • Start date
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1. Homework Statement
(a) Show that [itex]\sum \frac 1n[/itex] is not convergent by showing that the partial sums are not a Cauchy sequence
(b) Show that [itex]\sum \frac 1{n^2}[/itex] is convergent by showing that the partial sums form a Cauchy sequence


2. Homework Equations
Given epsilon>0, a sequence is Cauchy if there exists an N such that [itex]|a_m-a_n|<\epsilon[/itex] for every m,n>N.


3. The Attempt at a Solution
For part (a), the sequence terms are [itex]a_n=1+1/2+\ldots+1/n[/itex], so assuming that m>n,
[tex]|a_m-a_n|=1/m+1/(m-1)+\ldots+1/(n+1)<\frac{m-n}{n+1}<\frac{m-n}{n}[/tex].

Now, if I take epsilon=1/2 and suppose that m>n>N implies that the distance between two elements is less than epsilon. But m=2n>n>N gives the difference to be 1, which is greater than epsilon, so the sequence is not Cauchy. I think I've gotten this half - am I correct??

For part (b), I'm not sure how to do it. The sequence terms are [itex]a_n=1+1/2^2+\ldots+1/n^2[/itex], so again assuming that m>n, we have

[tex]|a_m-a_n|=1/m^2+1/(m-1)^2+\ldots+1/(n+1)^2<\frac{m-n}{(n+1)^2}<\frac{m-n}{n^2}<\frac{m}{n^2}[/tex]

I want to be able to find N such that m>n>N implies that this difference is less than epsilon, right? To do that, since I've assumed that m>n, if I can eliminate m from the expression, I'm good to go, but I can't figure out how to do it. Help!
 
Last edited:

Answers and Replies

62
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I'm grasping at straws here, but I want to show that [itex]m/n^2<\epsilon[/itex]. So I can say, since [itex]n<m[/itex], that [itex]n/n^2=1/n<m/n^2<\epsilon[/itex].

I know this probably isn't right, but if anybody could give me a hand it'd be appreciated.
 

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