What is the sum from n=1 to infinity of 2/(5^{n+2)} ?

The sum from n=1 to infinity of 2/(5^{n+2)} is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=1 to infinity of 2/(5^{n+2)}

n = 1 \sum \infty \frac{2}{5 ^{n + 2}}

converges

Solution steps

n = 1 \sum \infty \frac{2}{5 ^{n + 2}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{5 ^{n + 2}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

\frac{d}{d x} (\frac{e ^{x^{2}} + 3 e ^{x}}{e ^{x}})

\int 12 3 81 - 9 x - 36 d x

\int \frac{ln ( t )}{t ^{3}} d t

x \to 3 - lim (\frac{2}{x ^{2} - 9})

d er i v a t i v e e^{9 - x}

What is the sum from n=1 to infinity of 2/(5^{n+2)} ?
The sum from n=1 to infinity of 2/(5^{n+2)} is converges