What is the sum from k=3 to infinity of 1/(4k^2-1) ?

The sum from k=3 to infinity of 1/(4k^2-1) is 1/10

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sum from k=3 to infinity of 1/(4k^2-1)

k = 3 \sum \infty \frac{1}{4 k ^{2} - 1}

\frac{1}{10}

Solution steps

k = 3 \sum \infty \frac{1}{4 k ^{2} - 1}

Apply Telescoping Series Test:

\frac{1}{10}

= \frac{1}{10}

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 1}

n = 2 \sum \infty \frac{n !}{( 2 n ) !}

n = 1 \sum \infty \frac{1}{n ^{3} + n}

n = 1 \sum \infty \frac{1}{n ^{3} + 8}

n = 1 \sum \infty \frac{n}{n + 10}

What is the sum from k=3 to infinity of 1/(4k^2-1) ?
The sum from k=3 to infinity of 1/(4k^2-1) is 1/10