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

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

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

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

diverges

Solution steps

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

Apply log rule:

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

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

= n = 1 \sum \infty ln (n) - ln (n + 1)

Expand telescoping series:

diverges

= diverges

y^{^{''}} - 6 y^{^{'}} + 11 y = 0

n = 0 \sum \infty \frac{1}{n + 11}

y = sin^{2} (x)

\int 4 sin (4 x) d x

f (x) = \frac{- 1}{x}, at x = 16

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