How to prove divergence of harmonic series by eps-delta proof?

  • #1
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Homework Statement:

Let ##\{s_n\}_{n\in \mathbb{N}}=\{\sum_{1\leq k \leq n} \frac{1}{k}\}_{n\in \mathbb{N}}##. Show that this sequence cannot converge.

Relevant Equations:

A sequence ##\{a_n\}_{n\in \mathbb{N}}## is Cauchy (or equivalently, convergent) iff for all ##\epsilon >0##, there is a positive integer ##N## such that for all ##n,m\geq N##, ##|a_n-a_m|<\epsilon##.

Want to prove:
There is a ##\epsilon>0## such that for all positive integers ##N##, there is ##n,m\geq N## such that ##|s_n-s_m|\geq \epsilon##
Set ##\epsilon=\frac{1}{2}##. Let ##N\in \mathbb{N}## and choose ##n=N,m=2N##. Then:

##\begin{align*}
\left|s_N-s_{2N}\right|&=&\left|\sum_{l=1}^N \frac{1}{l} - \sum_{l=1}^{2N} \frac{1}{l}\right|\\
&=&\left|\left(1+\frac{1}{2}+...+\frac{1}{N}\right)-\left(1+...+\frac{1}{N}+...+\frac{1}{2N}\right)\right|\\
&=&\left|\frac{1}{N+1}+...+\frac{1}{N+N}\right|\\
&>&\left|\frac{1}{N+N}+...+\frac{1}{N+N}\right|=\left|\frac{N}{N+N}\right|\\
&=&\left|\frac{N}{2N}\right|=\frac{1}{2}=\epsilon
\end{align*}##
 
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Answers and Replies

  • #2
Delta2
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I think your work is correct. You prove that the sequence of partial sums is not Cauchy, and hence (by equivalence of the notions of Cauchy and Convergence for a sequence) is not convergent.
 
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  • #3
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Thanks for checking my work.
 
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