What is the sum from n=2 to infinity of 3/(5^{n-1)} ?

The sum from n=2 to infinity of 3/(5^{n-1)} 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=2 to infinity of 3/(5^{n-1)}

n = 2 \sum \infty \frac{3}{5 ^{n - 1}}

converges

Solution steps

n = 2 \sum \infty \frac{3}{5 ^{n - 1}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 2 \sum \infty \frac{1}{5 ^{n - 1}}

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 1 \sum \infty \frac{3}{1 + n ^{2}}

n = 1 \sum \infty 2^{(- 1)^{n}}

n = 1 \sum \infty cos (\frac{17}{n})

n = 0 \sum \infty \frac{1 ^{n}}{n + 1}

n = 2 \sum \infty \frac{n !}{n ^{n}}

What is the sum from n=2 to infinity of 3/(5^{n-1)} ?
The sum from n=2 to infinity of 3/(5^{n-1)} is converges