What is the sum from n=0 to infinity of 2(-5)^n ?

The sum from n=0 to infinity of 2(-5)^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=0 to infinity of 2(-5)^n

n = 0 \sum \infty 2 (- 5)^{n}

diverges

Solution steps

n = 0 \sum \infty 2 (- 5)^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 0 \sum \infty (- 5)^{n}

Apply Series Geometric Test:

diverges

= 2 diverges

= diverges

\frac{1}{s ^{3} - 7 s}

y^{^{'''}} + 64 y^{^{'}} = 0

f (x) = ln (100 sin^{2} (x))

\frac{d ^{2}}{d x ^{2}} (x^{\frac{8}{9}} + 6 x)

x \to - 1 lim (2 x + 4)

What is the sum from n=0 to infinity of 2(-5)^n ?
The sum from n=0 to infinity of 2(-5)^n is diverges