What is the sum from n=1 to infinity of (ln(n^2))/n ?

The sum from n=1 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=1 to infinity of (ln(n^2))/n

n = 1 \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 = 1 \sum \infty \frac{5}{n ^{2} + 6}

n = 100 \sum \infty \frac{1}{n}

n = 2 \sum \infty (- 0.4)^{n}

n = 0 \sum \infty π^{- 3 n}

n = 1 \sum \infty \frac{4 n}{n ^{2} + 1}

What is the sum from n=1 to infinity of (ln(n^2))/n ?
The sum from n=1 to infinity of (ln(n^2))/n is diverges