What is the sum from n=3 to infinity of 8/(n^2-1) ?

The sum from n=3 to infinity of 8/(n^2-1) is converges

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sum from n=3 to infinity of 8/(n^2-1)

n = 3 \sum \infty \frac{8}{n ^{2} - 1}

converges

Solution steps

n = 3 \sum \infty \frac{8}{n ^{2} - 1}

Apply the constant multiplication rule:

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

= 8 \cdot n = 3 \sum \infty \frac{1}{n ^{2} - 1}

Apply Series Integral Test:

converges

= 8 converges

= converges

4 x^{3} + 6 x^{2}

\frac{\partial}{\partial x} (6 x + 3 y - 7)

\frac{d}{d y} (2 e^{y})

x y + y^{^{'}} = 78 x

\int_{0}^{1} 2^{u} 2 u d u

What is the sum from n=3 to infinity of 8/(n^2-1) ?
The sum from n=3 to infinity of 8/(n^2-1) is converges