What is the sum from n=1 to infinity}log_{e of (n/(n+1)) ?

The sum from n=1 to infinity}log_{e of (n/(n+1)) is diverges

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sum from n=1 to infinity}log_{e of (n/(n+1))

n = 1 \sum \infty lo g_{e} (\frac{n}{n + 1})

diverges

Solution steps

n = 1 \sum \infty lo g_{e} (\frac{n}{n + 1})

Apply log rule:

lo g_{e} (x) = ln (x)

= 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

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What is the sum from n=1 to infinity}log_{e of (n/(n+1)) ?
The sum from n=1 to infinity}log_{e of (n/(n+1)) is diverges