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

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

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

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

diverges

Solution steps

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

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{k}{k + 1}) = ln (k) - ln (k + 1)

= k = 2 \sum \infty ln (k) - ln (k + 1)

Expand telescoping series:

diverges

= diverges

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

n = 5 \sum \infty e^{5 - 6 n}

n = 0 \sum \infty \frac{1}{n ^{10}}

n = 1 \sum \infty 2 (- \frac{1}{5})^{2 n}

n = 1 \sum \infty π

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