What is the sum from n=1 to infinity of 2-7/(8^n) ?

The sum from n=1 to infinity of 2-7/(8^n) is diverges

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 2-7/(8^n)

n = 1 \sum \infty 2 - \frac{7}{8 ^{n}}

diverges

Solution steps

n = 1 \sum \infty 2 - \frac{7}{8 ^{n}}

Apply Series Divergence Test:

diverges

= diverges

\int 3^{x + 1} d x

h \to 0 lim (\frac{1}{h})

9 x (\frac{d y}{d x}) = 2 x e^{x} - 9 y + 6 x^{2}

\int \frac{1}{1 + e ^{4 x}} d x

e^{- x}

What is the sum from n=1 to infinity of 2-7/(8^n) ?
The sum from n=1 to infinity of 2-7/(8^n) is diverges