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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 4 converges

= converges

x \to 0 lim (- 3 e^{3 x})

\int x e^{- \frac{x}{4}} d x

x + \frac{1}{x}, at x = 4

\int ln^{2} (c) os x tan (x)

\frac{d y}{d x} + y = x y^{3}

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