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This question already has an answer here:

There's a limit that I am unable to solve. I think it should be equal to $\infty$. $$\lim_{n\to\infty}\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$$

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marked as duplicate by Martin Sleziak, Dan Rust, Namaste, Najib Idrissi, Mark Bennet Apr 24 '14 at 14:15

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    $\begingroup$ Bound it below by an integral. $\endgroup$ – Qiaochu Yuan Apr 4 '11 at 15:33
  • $\begingroup$ upload.wikimedia.org/wikipedia/commons/0/04/Gamma-area.svg $\endgroup$ – JavaMan Apr 4 '11 at 16:49
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    $\begingroup$ This might be weird thing to ask...how do you know that $\log(x)$ diverges? $\endgroup$ – Mitch Apr 4 '11 at 16:50
  • $\begingroup$ @Mitch Because it has no upper bound, I guess. $\endgroup$ – Paul Manta Apr 4 '11 at 17:45
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    $\begingroup$ @Mitch: $\log(x)\to \infty$ as $x\to \infty$ because if you are given $M$, no matter how large, then $x>e^M$ implies $\log(x)>M$. $\endgroup$ – Jonas Meyer Apr 4 '11 at 18:52
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Write $$s_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}$$ and observe that $$s_{2n} - s_{n} = \frac{1}{n+1} + \frac{1}{n+2} + \cdots+ \frac{1}{2n} \geq \underbrace{\frac{1}{2n} + \cdots + \frac{1}{2n}}_{n\text{ terms}} = n \frac{1}{2n} = \frac{1}{2}$$ so that the sequence $(s_{n})_{n \in \mathbb{N}}$ can't be a Cauchy sequence, hence it can't converge. Put differently, $s_{2^{n}} \geq \frac{n}{2}$.


Edit:

What I was getting at seems not what Paul wanted to hear, but let me finish that argument nevertheless. It should be clear that $s_{n+1} = s_{n} + \frac{1}{n+1}$ so that $s_{n+1} \gt s_{n}$ or in words, the sequence $s_{n}$ is (strictly) monotonically increasing. I've argued that $s_{2^{n}} \geq \frac{n}{2}$ (see also Alexander's answer). Therefore we have $s_{k} \geq \frac{n}{2}$ for each $k \geq 2^{n}$ and as $\frac{n}{2}$ tends to infinity with $n$ so must $s_{k}$, as $k \to \infty$.

To find a divergent lower integral bound, note that the inequality $s_{n} \geq \int_{1}^{n} \frac{1}{x}\,dx = \log{n}$ follows from considering the upper Riemann sum associated to the partition $(1,2,\ldots,n)$ of $[1,n]$. Again this shows that $s_{n} \to \infty$ because $\log{n} \to \infty$ with $n \to \infty$ as Jonas explained in a comment to your question.

I hope that helps.

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  • $\begingroup$ I added in the integral trick because you seemed to have asked about it. I think the first argument I gave or @Moron's neat answer is the way to go. $\endgroup$ – t.b. Apr 4 '11 at 19:13
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Let $S_n = \sum_{k=1}^n \frac{1}{k}$, then $S_{2^{n+1}} - S_{2^n+1} \geq 2^n\frac{1}{2^{n+1}} = \frac{1}{2}$.

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If $\sum \dfrac{1}{n}$ were convergent to $S$, then it is absolutely convergent.

As a result we have that

$S_1 = \sum \dfrac{1}{2n-1} = 1 + \dfrac{1}{3} + \dfrac{1}{5} + \dots $

and

$S_2 = \sum \dfrac{1}{2n} = \dfrac{1}{2} + \dfrac{1}{4} + \dots$

are both absolutely convergent and

we have

$S_1 + S_2 = S$

and

$S_2 = S/2$ and thus

$S_1 = S_2$

which is not possible as $S_1 \gt S_2$.

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  • $\begingroup$ @Alexander Your objection does not apply. Each term in $S_1$ is greater than the corresponding term in $S_2$, hence, assuming (as Moron does) that $S$ converges, you see that $S_1$ is greater than $S_2$. $\endgroup$ – Did Apr 4 '11 at 21:20

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