What is the sum from n=1 to infinity of 5/(4n^2-1) ?

The sum from n=1 to infinity of 5/(4n^2-1) is 5/2

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sum from n=1 to infinity of 5/(4n^2-1)

n = 1 \sum \infty \frac{5}{4 n ^{2} - 1}

\frac{5}{2}

Solution steps

n = 1 \sum \infty \frac{5}{4 n ^{2} - 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} - 1}

Apply Telescoping Series Test:

\frac{1}{2}

= 5 \cdot \frac{1}{2}

Simplify

5 \cdot \frac{1}{2} : \frac{5}{2}

= \frac{5}{2}

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What is the sum from n=1 to infinity of 5/(4n^2-1) ?
The sum from n=1 to infinity of 5/(4n^2-1) is 5/2