What is the sum from n=6 to infinity of 5/(n^2-25) ?

The sum from n=6 to infinity of 5/(n^2-25) is converges

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sum from n=6 to infinity of 5/(n^2-25)

n = 6 \sum \infty \frac{5}{n ^{2} - 25}

converges

Solution steps

n = 6 \sum \infty \frac{5}{n ^{2} - 25}

Apply the constant multiplication rule:

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

= 5 \cdot n = 6 \sum \infty \frac{1}{n ^{2} - 25}

Apply Series Integral Test:

converges

= 5 converges

= converges

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\frac{d}{d x} (x^{3} + 3 x^{2} - 6 x - 8)

\int \frac{( arctan ( x ) ) ^{7}}{x ^{2} + 1} d x

What is the sum from n=6 to infinity of 5/(n^2-25) ?
The sum from n=6 to infinity of 5/(n^2-25) is converges