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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= 2 converges

= converges

n \to \infty lim (- 2 (n - 4))

n = 3 \sum \infty \frac{4 ^{n}}{5 ^{n}}

\frac{d}{d x} (a b^{x})

x^{3} (x - 1)^{2}

\int \frac{x + 1}{( x - 1 ) ^{3}} d x

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