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We have seen the harmonic series is a divergent series whose terms approach $0$. Show that $$\sum_{n = 1}^\infty \text{ln}\left(1 + \frac{1}{n}\right)$$ is another series with this property.

Denote $a_n = \text{ln}\left(1 + \frac{1}{n}\right).$ Then, $$\lim_{n \to \infty}\text{ln}\left(1 + \frac{1}{n}\right) = \text{ln}\left(1 + \lim_{n \to \infty}\frac{1}{n}\right) = 0,$$ since $\text{ln}(x)$ is a continues function on its domain. However, I can't seem to prove that the series is divergent. I know the proof for the harmonic series (which is quite clever in my opinion), and I wondered whether this question required a similar approach?

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    Can you show that $\sum \bigl(\frac{1}{n} - a_n\bigr)$ is (absolutely) convergent? – 2017-01-19
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    That series is a telescoping sum: $$\ln (1+ \frac{1}{n}) = \ln (n+1) - \ln n$$ – 2017-01-19
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    Another way to do it would be to use the inequality $\frac{x}{x+1} < \log \left( 1 + x \right)$ for $x>-1$. – 2017-01-19

3 Answers 3

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Hint:

$$\ln\frac21+\ln\frac32+\ln\frac43+\cdots\ln\frac{n+1}n=\ln\frac{2\cdot3\cdot4\cdot\cdots(n+1)}{1\cdot2\cdot3\cdot\cdots n}.$$

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    Haha, beautifully done. – 2017-01-19
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For all $n\in\mathbb{N}^\star$ :

$$\ln\left(1+\frac{1}{n}\right)=\ln\left(\frac{n+1}{n}\right)=\ln(n+1)-\ln(n)$$

so that :

$$\sum_{n=1}^N\ln\left(1+\frac{1}{n}\right)=\ln(N+1)-\ln(1)=\ln(N+1)\underset{N\to\infty}{\longrightarrow}+\infty$$

and the divergence of the series $\sum_{n\ge1}\ln\left(1+\frac{1}{n}\right)$ is proved.

Note that this gives us a proof (one of the easiest ones) of the divergence of the harmonic series, since :

$$\forall n\in\mathbb{N}^\star,\,\ln\left(1+\frac{1}{n}\right)\le\frac{1}{n}$$

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Hint:

This is a series with positive terms, and $\;\ln\Bigl(1+\dfrac1n\Bigr)\sim_\infty\dfrac1n$.