What is the sum from n=0 to infinity of (2n)/(n^2+1) ?

The sum from n=0 to infinity of (2n)/(n^2+1) 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=0 to infinity of (2n)/(n^2+1)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Limit Comparison Test:

diverges

= 2 diverges

= diverges

\frac{x}{x ^{2} + 4}

x x^{2} + 1

\frac{d}{d x} (\frac{x ^{2} + 1}{x})

\frac{d}{d x} (\frac{1}{x ^{3} + 2})

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

What is the sum from n=0 to infinity of (2n)/(n^2+1) ?
The sum from n=0 to infinity of (2n)/(n^2+1) is diverges