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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

x \to 0 + lim (x^{- \frac{17}{l n ( x )}})

\frac{d y}{d x} = 3 x (e^{2 y})

\frac{d}{d x} (\frac{3 x + 3}{4 e ^{x}})

\int \frac{2 x + 1}{x + 2} d x

\frac{d y}{d x} = (y + 5) (x - 2)

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