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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

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

n = 1 \sum \infty (cos (9))^{n}

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

n = 1 \sum \infty n^{4}

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

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