What is the sum from n=1 to infinity of 9ln(n/(n+1)) ?

The sum from n=1 to infinity of 9ln(n/(n+1)) is diverges

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sum from n=1 to infinity of 9ln(n/(n+1))

n = 1 \sum \infty 9 ln (\frac{n}{n + 1})

diverges

Solution steps

n = 1 \sum \infty 9 ln (\frac{n}{n + 1})

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 9 \cdot 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)

= 9 \cdot n = 1 \sum \infty ln (n) - ln (n + 1)

Expand telescoping series:

diverges

= diverges

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

n = 1 \sum \infty \frac{9}{n ^{\frac{3}{7}}}

n = 1 \sum \infty \frac{5}{n ^{3} + n}

n = 1 \sum \infty (- e)^{- n}

n = 0 \sum \infty (- \frac{5}{7})^{n}

What is the sum from n=1 to infinity of 9ln(n/(n+1)) ?
The sum from n=1 to infinity of 9ln(n/(n+1)) is diverges