What is the sum from n=2 to infinity of 3/(nln(n)) ?

The sum from n=2 to infinity of 3/(nln(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=2 to infinity of 3/(nln(n))

n = 2 \sum \infty \frac{3}{n ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{3}{n ln ( n )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot n = 2 \sum \infty \frac{1}{n ln ( n )}

Apply Series Integral Test:

diverges

= 3 diverges

= diverges

n = 1 \sum \infty \frac{2}{( 4 n ^{2} - 1 )}

n = 1 \sum \infty (e)^{- n}

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

n = 1 \sum \infty \frac{5 n}{5 n + 1}

k = 1 \sum \infty \frac{5}{k ( k + 2 )}

What is the sum from n=2 to infinity of 3/(nln(n)) ?
The sum from n=2 to infinity of 3/(nln(n)) is diverges