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

The sum from n=3 to infinity of 2/(n^2-2n) 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=3 to infinity of 2/(n^2-2n)

n = 3 \sum \infty \frac{2}{n ^{2} - 2 n}

converges

Solution steps

n = 3 \sum \infty \frac{2}{n ^{2} - 2 n}

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 2 converges

= converges

\frac{\partial}{\partial y} (- x - y)

x \to - \infty lim (e^{3 x})

y^{^{'}} - \frac{7}{x} y = \frac{y ^{4}}{x ^{5}}

x y d y + e^{- x} (y^{2} - 1) d x = 0

y^{^{'}} = \frac{( y - x )}{( x )}

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