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

The sum from n=3 to infinity of 3/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=3 to infinity of 3/n

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Cauchy's Convergence Condition:

diverges

= 3 diverges

= diverges

n = 1 \sum \infty \frac{7 n}{7 n + 1}

n = 1 \sum \infty 2 - \frac{1}{n}

n = 8 \sum \infty ln (\frac{n}{4})

k = 1 \sum \infty (cos (18))^{k}

n = 0 \sum \infty (sin (9))^{n}

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