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Prove that the sequence defined by partial sums of the Harmonic series

$$\left\{S_n\right\}^{\infty}_{n=1} =\left\{\sum_{k=1}^{n} \frac{1}{k}\right\}^{\infty}_{n=1} $$ is not a Cauchy sequence.

I have a theorem that states

Theorem: A sequence is Cauchy if and only if it is convergent.

I first need help in understanding the proof that the Harmonic series is divergent. I somewhat understand that since its a sum and not a sequence that the sum will obviously go to infinity.I however get confused with the "sequence defined by partial sums of the Harmonic series" part.

second I was wondering that after proving that the sequence is indeed divergent that I can use the theorem to say its not cauchy. How would I go about wording that?

user
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Ever Olivares
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  • Hi, it is not true in general that a sequence must diverge to infinity. For example consider the series $\sum_{n=1}^{\infty}\frac1{n^2}$, or the simpler $\sum_{n=1}^{\infty}\frac1{2^n}$. You will find both converge. –  Sep 29 '19 at 00:03
  • The sequence of partial sums is $(H_n)$, where $H_n=\sum_{k=1}^{n}\frac1{n}$ –  Sep 29 '19 at 00:07
  • For divergence of the series, have a look at https://math.stackexchange.com/q/255/418131 –  Sep 29 '19 at 00:14

4 Answers4

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It might be more instructive not to use the theorem at all.

You can prove directly that $S_n=\sum^n_{k=1}\frac{1}{k}$ is not Cauchy: if $n>m,$ we have $S_n-S_m=\frac{1}{m+1} + \frac{1}{m+2} +...+ \frac{1}{n} > \frac{n - m}{n} = 1 - m/n.$ Now, let $\epsilon=1/2.$ Then, if $n>2m,\ S_n-S_m> 1/2$ and so $(S_n)$ is not Cauchy.

Matematleta
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  • I choose to approach another way, but your answer seems intriguing enough to learn. I don't quite understand the approach , however I will ask one of my professors to help me understand it. Thank you! – Ever Olivares Sep 29 '19 at 18:11
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The wording is simple. Suppose, if possible, $(S_n)$ is Cauchy. Then, by the theorem, $S_n$ converges to some number $S$. By definition of convergence of a series this means that the series $\sum \frac 1 n$ converges to $S$. But this series is not covergent. Hence, $(S_n)$ cannot be Cauchy.

Kavi Rama Murthy
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The sequence defined is $\{H_1,H_2,\cdots,H_n\}$ that is the sequence of the harmonic numbers and we know that it is divergent.

Yes, since tha theorem states an "if and only if" condition, we can conlude that the sequence is not Cauchy.

user
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Cauchy sequences are bounded, while $H_n$ is not:

$$ H_n = \sum_{k=1}^{n}\frac{1}{k} > \sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right)=\sum_{k=1}^{n}\log(k+1)-\log(k) = \log(n).$$

Jack D'Aurizio
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