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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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You'd be interested in Arturo's answer here... – J. M. Oct 9 '11 at 15:02
@user825089: I've added LaTeX formatting to your question. Apologies if I changed your intended meaning in any way, please feel free to make further edits. – Zev Chonoles Oct 9 '11 at 15:54
@ZevChonoles Not at all, looks way better now! I can't do all this because I am new here. Thank you! – ivt Oct 9 '11 at 16:00
up vote 1 down vote accepted

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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Incredible! Yup, that's it! Thank you so much! – ivt Oct 9 '11 at 15:53
Can you tell me how do I make this "The Best Answer"? I am new here and it doesn't let me upvote... – ivt Oct 9 '11 at 15:59
Sorry, I'm new here too, so I'm not sure. I looks as though there should be an outline of a checkmark to the left of the answer, as shown here. – process91 Oct 9 '11 at 16:26
Done! Again thank you so much! – ivt Oct 9 '11 at 16:31 is weird. I tried to prove this formula inductively, but got some bad results. It appears that eighth term is not greater than 3. In fact, only the 11th term becomes greater than 3. I checked with Wolfram|Alpha. So, the formula is not correct. I will still leave it as a best answer because you were on the right track. However, I think I will need to re-ask this question... – ivt Oct 9 '11 at 21:12

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. – Jonas Teuwen Oct 9 '11 at 16:43

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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Thank you for the input. I knew about this Oresme trick. However, I don't have a series, I have a sequence of partial sums. So, this proof doesn't quite work here. What I need is a formula for finding the specific number N with respect to Epsilon. E.g. they give me e=10 and I say: OK here is the number N starting with which every term will be greater than epsilon. – ivt Oct 9 '11 at 15:22
Well, with this you can bound the partial sums as well? Just the ones for sum upper limit $2^n$ or something like that. Then you will have a divergent subsequence, hence the series will diverge. – Jonas Teuwen Oct 9 '11 at 15:23
Well, I cannot bound the partial sums, because the sequence diverges. In fact, I need to prove that it is unbounded in terms of E-N... – ivt Oct 9 '11 at 15:28
This argument will work. It is quite strange that you want to do this using the definition, that is unnecessarily complicated. – Jonas Teuwen Oct 9 '11 at 15:34
It's not that I want, I kind of have to. That proof is great, but it is not dealing with partial sums at all. It only deals with series. And I have partial sums. That proof wouldn't even make sense in terms of sequence, because there is no such thing as + in sequence. It is what it's called: a sequence of some numbers. – ivt Oct 9 '11 at 15:41

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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Welcome to math.stackexchange. Unfortunately, your first sentence is not true. Consider $\sum (-1)^n$. – Ragib Zaman Oct 9 '11 at 15:13
Sorry, that should be a series of positive terms. Thanks for the correction, I'll modify the answer. – process91 Oct 9 '11 at 15:19
Thank you! However, I need to prove the other way. You see, I am not interested in "iff" here. I'm interested in =>. So, if the sequence of partial sums is monotonic and unbounded => the series diverges. – ivt Oct 9 '11 at 15:26
Oh, in your original question you just said that you were tasked with proving that the sequence of partial sums diverges. – process91 Oct 9 '11 at 15:29
Well, iff is necessity and sufficiency. (<=>) Let's suppose you proved sufficiency with the integral method. But we still need to prove necessity in order to establish an iff in the first place. – ivt Oct 9 '11 at 15:32

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