What is the sum from n=5 to infinity of 6/(n^2-1) ?

The sum from n=5 to infinity of 6/(n^2-1) is converges

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sum from n=5 to infinity of 6/(n^2-1)

n = 5 \sum \infty \frac{6}{n ^{2} - 1}

converges

Solution steps

n = 5 \sum \infty \frac{6}{n ^{2} - 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 5 \sum \infty \frac{1}{n ^{2} - 1}

Apply Series Integral Test:

converges

= 6 converges

= converges

5^{x}

(- 4, - 4), (- 3, 0)

\frac{\partial}{\partial x} (18 x^{2} y + 12 x y^{2})

arcsin (\frac{x ^{2} - 1}{x})

x \to 3 lim (3 x + 9)

What is the sum from n=5 to infinity of 6/(n^2-1) ?
The sum from n=5 to infinity of 6/(n^2-1) is converges