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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

n = 0 \sum \infty \frac{4}{1 + 3 n ^{3}}

n = 0 \sum \infty \frac{n ^{2}}{5 + 2 n}

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

n = 1 \sum \infty \frac{1}{5 ^{n} - n}

i = 1 \sum \infty 5 \cdot 0. 2^{i - 1}

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