What is the sum from n=3 to infinity of (4^n)/(7^n) ?

The sum from n=3 to infinity of (4^n)/(7^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=3 to infinity of (4^n)/(7^n)

n = 3 \sum \infty \frac{4 ^{n}}{7 ^{n}}

converges

Solution steps

n = 3 \sum \infty \frac{4 ^{n}}{7 ^{n}}

n \to \infty lim (n (\frac{\frac{4 ^{n}}{7 ^{n}}}{\frac{4 ^{n + 1}}{7 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

\frac{d y}{d x} = 4 x^{3} y^{2}

\int x^{4} y^{3} d y

x = 9 y^{2}, x = 2 + 7 y^{2}

x^{- \frac{1}{2}}

\int_{- 1}^{1} (- x^{2} + 1)^{2} d x

What is the sum from n=3 to infinity of (4^n)/(7^n) ?
The sum from n=3 to infinity of (4^n)/(7^n) is converges