What is the sum from n=1 to infinity of 1/(4n^2-25) ?

The sum from n=1 to infinity of 1/(4n^2-25) is converges

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sum from n=1 to infinity of 1/(4n^2-25)

n = 1 \sum \infty \frac{1}{4 n ^{2} - 25}

converges

Solution steps

n = 1 \sum \infty \frac{1}{4 n ^{2} - 25}

Apply Series Integral Test:

converges

= converges

m = 2 \sum \infty \frac{2}{7 ^{m}}

n = 0 \sum \infty \frac{- 2}{n + 1}

n = 1 \sum \infty (- \frac{7}{9})^{n - 1}

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

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

What is the sum from n=1 to infinity of 1/(4n^2-25) ?
The sum from n=1 to infinity of 1/(4n^2-25) is converges