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This sum is convergence or divergence?

$$\sum_{n=1}^\infty \ln\left(\frac{n}{n+1}\right)$$

What I 've tried

  • Divergent test : inconclusive, limit is $0$
  • Comparison Test : inconclusive, larger term is divergent

$$\sum_{n=1}^\infty \ln\left(\frac{n}{n+1}\right) < \sum_{n=1}^\infty \frac{n}{n+1}$$

  • Limit Comparison Test : inconclusive, limit is $-1$ for $b_n = 1/n$
  • Ratio Test : inconclusive, using L'hopital's rule and the limit is $1$
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    $\begingroup$ Limit comparison test with the harmonic series is conclusive as you already have computed that the ratio of $\log(n/(n+1))$ and $1/n$ converges to $-1$. And on top of all, you may appeal to the direct computation of partial sums $$\sum_{k=1}^{n} \log\left(\frac{k}{k+1}\right)=-\log(n+1).$$ $\endgroup$ – Sangchul Lee Mar 4 '19 at 12:22
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This is a telescoping sum $\sum_{n\ge 1}(\ln n-\ln (n+1))=\ln 1-\ln\infty=-\infty$.

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Hint:

Try writing this sum for a particular $n$, and using properties of logarithms $\ln \frac{a}{b} = \ln a - \ln b$. Then think about limit.

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  • $\begingroup$ oh it's that easy doing this way. thank you $\endgroup$ – FIFATanathan Mar 4 '19 at 12:30

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