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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 2 converges

= converges

\frac{d P}{d t} = P - P^{2}

\frac{d y}{d x} = \frac{3}{x ^{2} y - 8 y}

\frac{\partial}{\partial x} (R (x)^{2} + x^{2} - x)

f (0) = (x^{4} + 3 x) (4 x^{5} + 4 x - 2)

- \frac{2}{x ^{2}}

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