What is the sum from n=2 to infinity of 1/(n^2+5n+6) ?

The sum from n=2 to infinity of 1/(n^2+5n+6) is 1/4

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sum from n=2 to infinity of 1/(n^2+5n+6)

n = 2 \sum \infty \frac{1}{n ^{2} + 5 n + 6}

\frac{1}{4}

Solution steps

n = 2 \sum \infty \frac{1}{n ^{2} + 5 n + 6}

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

n = 1 \sum \infty \frac{5 n}{3 n - 1}

\frac{d}{d x} (- \frac{2 x}{x - 1})

f (x) = 7 (2 - 3 x)^{4}

\int \frac{( x ^{am} - x ^{n} )}{x} d x

y^{^{'}} + 3 y = 2 x e^{- 3 x}

What is the sum from n=2 to infinity of 1/(n^2+5n+6) ?
The sum from n=2 to infinity of 1/(n^2+5n+6) is 1/4