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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\frac{d}{d x} (2 x (3 x^{2} - 7)^{3})

x \to 3 lim (\frac{x ^{2} - 1}{x - 3})

\int \frac{z}{z - 1} d z

\int \frac{2}{x ^{2} - x} d x

\frac{d v}{d t}

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