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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

\int \frac{cos ( 3 x ) cos ( 2 x )}{3} d x

\int_{0}^{\frac{π}{6}} 20 sec^{4} (x) d x

\int \frac{8 x + 6}{2 x ^{2} + 3 x - 2} d x

\int tan (3 x) d x

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

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