What is the sum from n=0 to infinity of ln(n/(n+1)) ?

The sum from n=0 to infinity of ln(n/(n+1)) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of ln(n/(n+1))

n = 0 \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

n = 1 \sum \infty \frac{1}{n + 2} - \frac{1}{n}

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

n = 0 \sum \infty \frac{1}{2 n - 1}

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

n = 1 \sum \infty 2^{n} + i

What is the sum from n=0 to infinity of ln(n/(n+1)) ?
The sum from n=0 to infinity of ln(n/(n+1)) is diverges