What is the sum from n=0 to infinity of 1/(n^2+9n+20) ?

The sum from n=0 to infinity of 1/(n^2+9n+20) is 1/4

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sum from n=0 to infinity of 1/(n^2+9n+20)

n = 0 \sum \infty \frac{1}{n ^{2} + 9 n + 20}

\frac{1}{4}

Solution steps

n = 0 \sum \infty \frac{1}{n ^{2} + 9 n + 20}

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

f (x) = x^{2} - 2 x + 3

y = \frac{- 8 x}{x ^{2} + 1}

\frac{4 x}{x + 3}

x \to - 1 lim (ln (5 + 2 x))

x \to 0 lim (e^{\frac{5}{x}})

What is the sum from n=0 to infinity of 1/(n^2+9n+20) ?
The sum from n=0 to infinity of 1/(n^2+9n+20) is 1/4