What is the sum from n=0 to infinity of (ln(n^2))/n ?

The sum from n=0 to infinity of (ln(n^2))/n 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^2))/n

n = 0 \sum \infty \frac{ln ( n ^{2} )}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{ln ( n ^{2} )}{n}

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty \frac{3}{3 + n ^{2}}

n = 1 \sum \infty \frac{n}{n ^{2} + 8}

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

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

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

What is the sum from n=0 to infinity of (ln(n^2))/n ?
The sum from n=0 to infinity of (ln(n^2))/n is diverges