# How do you determine if the series ln(1/2) + ln(1/3) + ln(3/4) + ... +ln[k/(k + 1)] + .... converges?

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3
Dec 2, 2017

First of all I think there is a little mistake, the second term would be:

$\ln \left(\frac{2}{3}\right)$ instead of $\ln \left(\frac{1}{3}\right)$.

So it would be:

${\sum}_{k = 1}^{+ \infty} \ln \left(\frac{k}{k + 1}\right)$.

We can use an important logarithmic rule that says:

$\ln \left(\frac{a}{b}\right) = \ln a - \ln b$.

So:

${\sum}_{k = 1}^{+ \infty} \ln \left(\frac{k}{k + 1}\right) = \ln \left(\frac{1}{2}\right) + \ln \left(\frac{2}{3}\right) + \ln \left(\frac{3}{4}\right) + \ldots =$

$= \ln 1 - \ln 2 + \ln 2 - \ln 3 + \ln 3 - \ln 4 + \ldots + \ln k - \ln \left(k + 1\right) + \ldots$.

The first term is zero, all the other will cancel each other, but there is always another that survive. The bigger becomes $k$ the bigger becomes $\ln \left(k + 1\right)$, the smaller becomes $- \ln \left(k + 1\right)$. So the series will diverge to $- \infty$.

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