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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 4 converges

= converges

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

n = 0 \sum \infty \frac{i ^{2 n}}{4 n}

n = 1 \sum \infty n (\frac{1}{7})^{n}

n = 1 \sum \infty \frac{( 3 n - 2 )}{n}

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

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