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

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

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 (5n)/(n^2+2)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

diverges

= 5 diverges

= diverges

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