What is the sum from n=1 to infinity of 4/(nln(n)) ?

The sum from n=1 to infinity of 4/(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=1 to infinity of 4/(nln(n))

n = 1 \sum \infty \frac{4}{n ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{4}{n ln ( n )}

Apply the constant multiplication rule:

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

= 4 \cdot n = 2 \sum \infty \frac{1}{n ln ( n )}

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

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

n = 1 \sum \infty \frac{5}{3 n ^{2}}

n = 1 \sum \infty \frac{1}{2 \cdot n - 1}

n = 1 \sum \infty \frac{1}{3 n ^{3} + 8}

n = 1 \sum \infty e^{- \frac{n}{9}}

What is the sum from n=1 to infinity of 4/(nln(n)) ?
The sum from n=1 to infinity of 4/(nln(n)) is diverges