What is the sum from n=8 to infinity of 1/(ln(9^n)) ?

The sum from n=8 to infinity of 1/(ln(9^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=8 to infinity of 1/(ln(9^n))

n = 8 \sum \infty \frac{1}{ln ( 9 ^{n} )}

diverges

Solution steps

n = 8 \sum \infty \frac{1}{ln ( 9 ^{n} )}

Simplify

\frac{1}{ln ( 9 ^{n} )} : \frac{1}{2 ln ( 3 ) n}

= n = 8 \sum \infty \frac{1}{2 ln ( 3 ) n}

Apply the constant multiplication rule:

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

= \frac{1}{2 ln ( 3 )} \cdot n = 8 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{1}{2 ln ( 3 )} diverges

= diverges

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

n = 0 \sum \infty \frac{1000}{n}

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

n = 0 \sum \infty \frac{e}{n}

n = 0 \sum \infty \frac{7}{π ^{n}}

What is the sum from n=8 to infinity of 1/(ln(9^n)) ?
The sum from n=8 to infinity of 1/(ln(9^n)) is diverges