What is the sum from n=5 to infinity of (7^n)/(8^n) ?

The sum from n=5 to infinity of (7^n)/(8^n) 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=5 to infinity of (7^n)/(8^n)

n = 5 \sum \infty \frac{7 ^{n}}{8 ^{n}}

converges

Solution steps

n = 5 \sum \infty \frac{7 ^{n}}{8 ^{n}}

n \to \infty lim (n (\frac{\frac{7 ^{n}}{8 ^{n}}}{\frac{7 ^{n + 1}}{8 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

\frac{d y}{d x} = 7^{x} ln (7) + \frac{2}{x ^{2} + 1}

\frac{d ^{3}}{d x ^{3}} (\frac{10}{x ^{2}})

\int (- e^{- x}) d x

\frac{d}{d x} (\frac{x ^{2} - 4}{x ^{2} - 3})

i = 0 \sum \infty (\frac{1}{3})^{i}

What is the sum from n=5 to infinity of (7^n)/(8^n) ?
The sum from n=5 to infinity of (7^n)/(8^n) is converges