What is the sum from n=1 to infinity of 3ln(n/(n+1)) ?

The sum from n=1 to infinity of 3ln(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=1 to infinity of 3ln(n/(n+1))

n = 1 \sum \infty 3 ln (\frac{n}{n + 1})

diverges

Solution steps

n = 1 \sum \infty 3 ln (\frac{n}{n + 1})

Apply the constant multiplication rule:

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

= 3 \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)

= 3 \cdot n = 1 \sum \infty ln (n) - ln (n + 1)

Expand telescoping series:

diverges

= diverges

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

n = 1 \sum \infty \frac{1}{n ^{1.2}}

n = 2 \sum \infty (- 0.15)^{n}

n = 0 \sum \infty \frac{20}{4 ^{n}}

n = 1 \sum \infty (\frac{π}{3})^{n}

What is the sum from n=1 to infinity of 3ln(n/(n+1)) ?
The sum from n=1 to infinity of 3ln(n/(n+1)) is diverges