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

\int \frac{x}{5 x - 2} d x

\int_{\frac{1}{3}}^{7} 6 x ln (3 x) d x

\frac{d}{d x} (\frac{4 x ^{3} + 2 x + 5}{x})

y^{^{'}} sin (y) = x^{4}

x \to 3 lim (6 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