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

In summary, for part (a), it was shown that the partial sums of the sequence \sum \frac 1n are not a Cauchy sequence, thus the series is not convergent. For part (b), it was shown that the partial sums of the sequence \sum \frac 1{n^2} form a Cauchy sequence, thus the series is convergent. The proof involved using the definition of a Cauchy sequence and choosing an appropriate epsilon value to show convergence.
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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

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.

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!
 
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  • #2
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.
 

1. What is the Divergent Harmonic Series?

The Divergent Harmonic Series is a mathematical series that diverges, meaning it has no finite sum. It is defined as the sum of the reciprocals of all positive integers, starting from 1. This series is represented by Hn = 1 + 1/2 + 1/3 + 1/4 + ... + 1/n.

2. What is a Convergent P-Series?

A Convergent P-Series is a mathematical series that converges, meaning it has a finite sum. It is defined as the sum of the reciprocals of all positive integers raised to a power, denoted by the letter p. This series is represented by Pn = 1 + 1/2p + 1/3p + 1/4p + ... + 1/np.

3. What is a Cauchy Sequence?

A Cauchy Sequence is a sequence of numbers that becomes arbitrarily close to each other as the sequence progresses. In other words, for any given tolerance, there exists a point in the sequence where all subsequent terms are within that tolerance of each other. This concept is important in the study of convergence and divergence of mathematical series.

4. How is the Divergent Harmonic Series related to the Convergent P-Series?

The Divergent Harmonic Series and Convergent P-Series are related through the value of the exponent, p. If p is greater than 1, the P-Series will converge, while if p is less than or equal to 1, the P-Series will diverge. This means that the Divergent Harmonic Series is a special case of the Convergent P-Series, where p = 1.

5. Why is the Divergent Harmonic Series considered to be a divergent series?

The Divergent Harmonic Series is considered to be a divergent series because the sum of its terms does not approach a finite value as the number of terms increases. In fact, as the number of terms increases, the sum gets larger and larger. This can be proven using the Cauchy Condensation Test, which shows that the series is equivalent to 1 + 1 + 1 + ... which clearly diverges.

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