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

The sum from n=1 to infinity of 1/(n^2+5n+6) is 1/3

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

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

\frac{1}{3}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{3}

= \frac{1}{3}

\frac{d}{d x} (4 + ln (x))

y^{^{'}} = y (x y^{6} + 1)

\int_{0}^{17} 2 π x (17 x - x^{2}) d x

x \to a lim (\frac{sec ( x ) - sec ( a )}{x - a})

\int (\frac{7}{x} + 7 x) d x

What is the sum from n=1 to infinity of 1/(n^2+5n+6) ?
The sum from n=1 to infinity of 1/(n^2+5n+6) is 1/3