What is the sum from n=1 to infinity of 2/(n^2+4n+3) ?

The sum from n=1 to infinity of 2/(n^2+4n+3) 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 2/(n^2+4n+3)

n = 1 \sum \infty \frac{2}{n ^{2} + 4 n + 3}

converges

Solution steps

n = 1 \sum \infty \frac{2}{n ^{2} + 4 n + 3}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

Apply Series Integral Test:

converges

= 2 converges

= converges

\frac{\partial}{\partial x} (x^{4} + 2 y^{6} - x y - x + 2)

\int_{0}^{4} π (x + 1)^{2} d x

f (x) = x^{3} + 2 x - 3

x^{2} + y^{2} = 50, (7, - 1)

x \to + (- 7) lim (- 13)

What is the sum from n=1 to infinity of 2/(n^2+4n+3) ?
The sum from n=1 to infinity of 2/(n^2+4n+3) is converges