What is the sum from n=1 to infinity of n/(3n) ?

The sum from n=1 to infinity of n/(3n) 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 n/(3n)

n = 1 \sum \infty \frac{n}{3 n}

diverges

Solution steps

n = 1 \sum \infty \frac{n}{3 n}

Apply the constant multiplication rule:

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

= \frac{1}{3} \cdot n = 1 \sum \infty \frac{n}{n}

Simplify

= \frac{1}{3} \cdot n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= \frac{1}{3} diverges

= diverges

n = 0 \sum \infty \frac{n}{2 n ^{3} + 1}

n = 0 \sum \infty n

n = 1 \sum \infty (\frac{5 π}{6})^{n}

k = 1 \sum \infty \frac{5}{k}

n = 0 \sum \infty \frac{1}{6 - n}

What is the sum from n=1 to infinity of n/(3n) ?
The sum from n=1 to infinity of n/(3n) is diverges