What is the sum from n=2 to infinity of ln((n+1)/n) ?

The sum from n=2 to infinity of ln((n+1)/n) is diverges

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sum from n=2 to infinity of ln((n+1)/n)

n = 2 \sum \infty ln (\frac{n + 1}{n})

diverges

Solution steps

n = 2 \sum \infty ln (\frac{n + 1}{n})

Expand

\frac{n + 1}{n} : 1 + \frac{1}{n}

= n = 2 \sum \infty ln (1 + \frac{1}{n})

Apply Series Limit Comparison Test:

diverges

= diverges

k = 1 \sum \infty \frac{3 k - 2}{3 ^{k}}

n = 1 \sum \infty \frac{1}{6 n ^{4} + 24 n}

n = 0 \sum \infty \frac{n ^{3}}{2 + n}

n = 1 \sum \infty \frac{1}{n ( 2 n + 1 )}

n = 1 \sum \infty \frac{1}{7 n + 3}

What is the sum from n=2 to infinity of ln((n+1)/n) ?
The sum from n=2 to infinity of ln((n+1)/n) is diverges