What is the sum from n=5 to infinity of (6^n)/(7^n) ?

The sum from n=5 to infinity of (6^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=5 to infinity of (6^n)/(7^n)

n = 5 \sum \infty \frac{6 ^{n}}{7 ^{n}}

converges

Solution steps

n = 5 \sum \infty \frac{6 ^{n}}{7 ^{n}}

n \to \infty lim (n (\frac{\frac{6 ^{n}}{7 ^{n}}}{\frac{6 ^{n + 1}}{7 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

x \to 0 + lim ((5 x)^{4 x})

f (x) = x cos (x^{2})

\int \frac{2}{2 x ^{2} + 4 x - 6} d x

\int e^{x^{3}} 6 x^{2} d x

\int \frac{2 x + 3}{9 - 4 x ^{2}} d x

What is the sum from n=5 to infinity of (6^n)/(7^n) ?
The sum from n=5 to infinity of (6^n)/(7^n) is converges