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I am trying to find whether $\sum_{n=1}^{\infty}\ln\left(\frac{n+2}{n+1}\right)$ converges or diverges. I used the limit test, and it comes out as inconclusive since $\lim_{n\rightarrow\infty}\ln\left(\frac{n+2}{n+1}\right) = 0$. When I put it into wolfram, it states the series diverges by comparison test. But I don't know how to set up the comparison test (what series to compare it to). All help in solving this would be greatly appreciated, thanks.

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5 Answers 5

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Hints:

$\ln\frac{n+2}{n+1}=\ln(n+2)-\ln(n+1)$.

Then $$\sum_{n=1}^{\infty}\ln\frac{n+2}{n+1}=\lim_{m\to \infty}\sum_{n=1}^{m}\ln\frac{n+2}{n+1}=\lim_{m\to \infty} [\ln(m+2)-\ln2]=+\infty$$

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  • $\begingroup$ Ahhh, I misread and I made the assumption that it converges, when in reality it diverges, so I tried to prove myself otherwise. I appreciate the help. $\endgroup$
    – Kenshin
    Oct 24, 2014 at 2:16
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Observe that: $\displaystyle \lim_{n \to \infty} \dfrac{\ln\left(1+\dfrac{1}{n+1}\right)}{\dfrac{1}{n+1}} = 1$, and the harmonic series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n+1}$ diverges, so the series $\displaystyle \sum_{n=1}^\infty \ln\left(1+\dfrac{1}{n+1}\right)$ also diverges.

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  • $\begingroup$ Thanks, another great way to prove divergence. $\endgroup$
    – Kenshin
    Oct 24, 2014 at 3:16
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Since $\log \frac {n + 2} {n + 1} = \log (1 + \frac {1} {n + 1}) \sim \frac {1} {n +1}$, we have

$$\sum_{n = 1}^{\infty} \log \frac {n + 2} {n + 1} \sim \sum_{n = 1}^{\infty} \frac {1} {n +1}$$

which diverges.

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A cool way think about it is this-

$\sum_{n=1}^{\infty}\ln\left(\frac{n+2}{n+1}\right)=\ln\left(\Pi_{n=1}^{\infty}\left(\frac{n+2}{n+1}\right)\right)=\ln\left(\lim_{N\rightarrow\infty}\Pi_{n=1}^{N}\left(\frac{n+2}{n+1}\right)\right)=\ln\left(\lim_{N\rightarrow\infty}\left(\frac 3 2\cdot\frac 4 3\cdot\dotso\cdot\frac{N+1}N\cdot\frac{N+2}{N+1}\right)\right)=\ln\left(\lim_{N\rightarrow\infty}\frac{N+2}2\right)=\infty$


In general, when you have an infinite sum of $\ln$s, it is sometimes beneficial to look at it as $\ln$ of the product.

$\sum_{n=1}^{\infty}\ln\left(a_n\right)=\ln\left(\Pi_{n=1}^{\infty}\left(a_n\right)\right)$

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  • $\begingroup$ How is this different from Paul's answer? $\endgroup$
    – mrf
    Sep 25, 2015 at 22:23
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    $\begingroup$ You are right, now I see it has the same idea. But I think my answer gives intuition and explains it clearly for a beginner like myself, giving a method for future similar problems. $\endgroup$
    – Whyka
    Sep 25, 2015 at 22:32
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We know that $\ln(1+x)\leq x$ for all $x>-1$ which gives $-\ln(1+x)\geq -x$. Therefore we have that, for all $n\in\mathbb{N}$, $$\ln\left(\frac{n+2}{n+1}\right)=-\ln\left(\frac{n+2-1}{n+2}\right)$$ $$\implies\ln\left(\frac{n+2}{n+1}\right)=-\ln\left(1-\frac{1}{n+2}\right)\geq \frac{1}{n+2}$$ But the series $\sum_{n=1}^\infty \frac{1}{n+2}$ diverges.

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