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Show that $$\sum\limits_{k=1}^n \frac{1}{k} \geq \log(n)$$ Use this to deduce that the series $$\sum\limits_{k=1}^\infty \frac{1}{k}$$ diverges. Hint: Use the estimate $$\frac{1}{k} \geq \int_k^{k+1} \frac{1}{x}\,\mathrm{d}x$$

Note that all instances of "$\log$" mean the natural logarithm, the inverse of the exponential. After doing the integration, I've concluded that $\frac{1}{k} \geq \log(1+\frac{1}{k})>-\log(k)$

I suspect I'm supposed to use the hint to arrive at the first inequality, and then since $\log(n)$ can become arbitrarily large as $n$ increases, so must the harmonic series (essentially the comparison test).

I'm having trouble making this jump. Hints are appreciated, but please no solutions.

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$$\frac1k\ge\int_k^{k+1}\frac{dx}x=\log\frac{k+1}k\implies\sum_{k=1}^n\frac1k\ge\log\left(\prod_{k=1}^n\frac{k+1}k\right)\ge\log n$$

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  • $\begingroup$ I'm with you up until the last equality - I keep getting $\log(n+1)$. Not that it matters, since $\log(n+1)>\log(n)$, so the conclusion I need can be reached regardless. $\endgroup$ – Ducky Feb 2 '15 at 21:40
  • $\begingroup$ Yes, you get $\;\log(n+1)\;$ indeed...and indeed it doesn't matter for the reason you mention. Editing the typo. $\endgroup$ – Timbuc Feb 2 '15 at 21:41
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we can use the fact that $$\dfrac{1}{k-1} + \dfrac{1}{k}+\dfrac{1}{k+1} = \dfrac{1}{k}+\dfrac{2k}{k^2 - 1} > \dfrac{1}{k}+\dfrac{2}{k} = \dfrac{3}{k}\text{ for } k \ge 2$$ to show that $s=1 + \frac12 + \frac13 + \cdots$ cannot be finite, therefore the harmonic series must diverge.

$\begin{align} s &= \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} +\dfrac{1}{4} + \dfrac{1}{5}+ \cdots \\ &=1+\left(\dfrac{1}{2} + \dfrac{1}{3}+\dfrac{1}{4} \right)+ \left(\dfrac{1}{5} + \dfrac{1}{6}+\dfrac{1}{7} \right) + \left(\dfrac{1}{8} + \dfrac{1}{9}+\dfrac{1}{10} \right) + \left(\dfrac{1}{11} + \dfrac{1}{12}+\dfrac{1}{13} \right) + \cdots \\ &> 1+3\left(\frac13 + \frac16+\frac19+\frac1{12} \cdots\right) = 1+1 +\frac12 + \frac13+\frac14 + \cdots\\ &=1 + s \end{align}$

we have $ > s+1$, therefore $s$ cannot be a finite number.

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  • $\begingroup$ Why is $s>s-\frac16$ a contradiction? $\endgroup$ – LutzL Feb 2 '15 at 23:02
  • $\begingroup$ @LutzL, thanks for spotting the error. i have fixed the problem. $\endgroup$ – abel Feb 3 '15 at 16:28
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An other way:

Let $S_n=\sum_{k=1}^n\frac{1}{k}$.

$$|S_{2n}-S_n|=\sum_{k=n+1}^{2n}\frac{1}{k}=\underbrace{\frac{1}{n+1}}_{\geq \frac{1}{2n}}+\underbrace{\frac{1}{n+2}}_{\geq \frac{1}{2n}}+...+\underbrace{\frac{1}{2n}}_{\geq \frac{1}{2n}}\geq \frac{n}{2n}=\frac{1}{2}$$

and thus, $(S_n)$ is not a Cauchy sequence, therefore it doesn't converge.

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HINT:$$\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})...\gt\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})\cdots$$

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    $\begingroup$ Did you read the question, or just the title? $\endgroup$ – Akiva Weinberger Jul 21 '15 at 3:06

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