What is the sum from n=0 to infinity of 1/(n^2+3n+2) ?

The sum from n=0 to infinity of 1/(n^2+3n+2) is 1

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sum from n=0 to infinity of 1/(n^2+3n+2)

n = 0 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

1

Solution steps

n = 0 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

Apply Telescoping Series Test:

1

= 1

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

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

(0, - 4), (1, 2)

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

(1 - 5 y + x) \frac{d y}{d x} + y = 5 x - 1

What is the sum from n=0 to infinity of 1/(n^2+3n+2) ?
The sum from n=0 to infinity of 1/(n^2+3n+2) is 1