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I've been tasked to prove that the sequence of partial sums of harmonic series diverges.

$\lim_{n\to\infty}\;\;\sum_{i=1}^n \frac{1}{i}=\infty$

I decided to prove this using the Monotone convergence theorem. I managed to prove that the sequence in monotonic using induction. However, I don't know how to prove that it is unbounded.

Here's an example how to do this for:

$X_n = n$

$\lim_{n\to\infty} n = \infty $

$∀ε>0,\;\; ∃N(ε)∈\mathbb{N} : ∀n > N(ε), |{X_n}| > ε$

$n > ε$
$N(ε) = \lfloor ε\rfloor + 1$

So for every epsilon we can find a number $N(ε)$ that for every number greater than this number the $n$-term of a sequence will be greater than epsilon.

However, I have no idea which $N$ with respect to epsilon should I take in the case of partial sums, because there is that $∑$ sign.

Thank you so much!

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    @ZevChonoles Not at all, looks way better now! I can't do all this because I am new here. Thank you!2011-10-09

4 Answers 4

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If you can prove that a series of positive terms diverges, you know that the sequences of partial sums is unbounded (there's an iff relationship between them). There are many ways to prove that this series diverges, but a simple one is to use the integral test, which says that, if $f(k)>0$ for all $k$, the sequence $s_n=\sum_{k=1}^n f(k)$ converges iff the sequence $t_n=\int_1^n f(x) dx$ converges. Taking $f(x)=\frac 1 x$, you should see that $t_n$ diverges...

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    You are probably right, it could be an iff statement. However, we haven't learned neither derivatives nor integrals yet, so I simply cannot use that...I am even starting to wonder if that it is impossible to prove using E-N and I have to wait until we learn integrals.2011-10-09
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One can use the comparison test.

$1 + \frac12 + \frac13 + \frac14 + \ldots > 1 + \frac12 + \frac14 + \frac14 + \ldots$

So, in the RHS we can group the terms that add up to one. Take limits and there you go.

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    Yes, now I starting to realize what you meant by Sn, i$f$ we combine this with **process9$1$* formula, we get S2n = 2^(n-1)2011-10-09
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Here's one way to show the harmonic series diverges to $\infty$: $ \begin{align} & & 1 & {}+ \frac13 + \frac15 + \frac17 + \frac19 + \frac{1}{11} + \cdots \\ \\ & & & {}+ \frac12 + \frac14 + \frac16 + \frac18 + \frac{1}{10} + \cdots \\ \\ & & \ge 1 & {}+ \frac12 + \frac14 + \frac16 + \frac18 + \frac{1}{10} + \cdots \\ \\ & & & {}+ \frac12 + \frac14 + \frac16 + \frac18 + \frac{1}{10} + \cdots \\ \\ \\ & & = 1 & {} + 1 + \frac12+\frac13+\frac14+\frac15+\cdots \end{align} $ So if the sum $S$ is a finite number then $ S \ge 1+S. $

The OP wants to phrase this in terms of finite partial sums explicitly rather than implicitly as in cases like this. So look at $ \sum_{n=1}^N \frac 1n $ where $N$ odd. $ \begin{align} \sum_{n=1}^N \frac 1n & = 1 + \sum_{\text{odd }n\in\{1,\ldots, N\}} \frac 1n + \sum_{\text{even }\ n\in \{2,\ldots,N-1\}} \frac 1n \\ \\ & \ge 1 + \sum_{\text{even }\ n\in \{2,\ldots,N-1\}} \frac 1n + \sum_{\text{even }\ n\in \{2,\ldots,N-1\}} \frac 1n \\ \\ & = 1 + \sum_{n=1}^{(N-1)/2} \frac 1n. \end{align} $ Then show that the two sums in $ \sum_{n=1}^N \frac 1n \ge 1 + \sum_{n=1}^{(N-1)/2} \frac 1n $ must both approach the same limit $S$ as $N\to\infty$.

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    This is a cute argument.2011-10-09
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If you are set on using the $\epsilon, N$ method, then you can note that the first partial sum is greater than 1, the second is greater than $\frac 1 2$, the fourth is greater than $1+2 \cdot \frac 1 2$, the eighth is greater than $1+3 \cdot \frac 1 2$, and so forth. Inductively, the $2^{\epsilon}$th partial sum is greater than $1+\epsilon \cdot \frac 1 2$ where $\epsilon\in \mathbb N$. This should lead you to the value for $N$.

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    OK, I fixed it now.2011-10-10