What is the sum from n=2 to infinity of (9n)/(ln(n)) ?

The sum from n=2 to infinity of (9n)/(ln(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 (9n)/(ln(n))

n = 2 \sum \infty \frac{9 n}{ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{9 n}{ln ( n )}

Apply the constant multiplication rule:

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

= 9 \cdot n = 2 \sum \infty \frac{n}{ln ( n )}

Apply Series Divergence Test:

diverges

= 9 diverges

= diverges

k = 1 \sum \infty [(\frac{k}{2 ^{k}})]

k = 0 \sum \infty - 1^{k}

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

n = 0 \sum \infty e^{- 1 n}

n = 1 \sum \infty \frac{2 n}{7 n - 6}

What is the sum from n=2 to infinity of (9n)/(ln(n)) ?
The sum from n=2 to infinity of (9n)/(ln(n)) is diverges