What is the sum from n=1 to infinity of 1/(3^n-2) ?

The sum from n=1 to infinity of 1/(3^n-2) 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=1 to infinity of 1/(3^n-2)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

k = 0 \sum \infty (0.25)^{k}

n = 0 \sum \infty (n - 1)^{2}

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

n = 0 \sum \infty \frac{1}{2 n ^{2}}

n = 1 \sum \infty \frac{6 ^{n}}{n ^{k}}

What is the sum from n=1 to infinity of 1/(3^n-2) ?
The sum from n=1 to infinity of 1/(3^n-2) is converges