What is the sum from n=1 to infinity}(((-7)^{n-1 of))/(8^n) ?

The sum from n=1 to infinity}(((-7)^{n-1 of))/(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=1 to infinity}(((-7)^{n-1 of))/(8^n)

n = 1 \sum \infty \frac{( ( - 7 ) ^{n - 1} )}{8 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{( - 7 ) ^{n - 1}}{8 ^{n}}

Apply Series Ratio Test:

converges

= converges

y^{^{''}} - 14 y^{^{'}} + 49 y = t^{- 2} e^{7 t}

x \to - \infty lim (\frac{7}{x} - 4)

\int \frac{1}{w ^{3}} d w

12 x + 16 x^{\frac{3}{4}}

f (x) = x^{2} - 4, (- 2, 3)

What is the sum from n=1 to infinity}(((-7)^{n-1 of))/(8^n) ?
The sum from n=1 to infinity}(((-7)^{n-1 of))/(8^n) is converges