# $\sum_{n=1}^\infty \ln\left(\frac{n}{n+1}\right)$ convergence or divergence [duplicate]

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 $$1$$
• 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 $$\sum_{k=1}^{n} \log\left(\frac{k}{k+1}\right)=-\log(n+1).$$ – Sangchul Lee Mar 4 '19 at 12:22

This is a telescoping sum $$\sum_{n\ge 1}(\ln n-\ln (n+1))=\ln 1-\ln\infty=-\infty$$.
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.