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

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

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

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

\frac{1}{2}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

n = 2 \sum \infty \frac{10}{ln ( n )}

n = 0 \sum \infty \frac{9}{2} (\frac{1}{2})^{n}

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

k = 0 \sum \infty (\frac{1}{2})^{k}

n = 0 \sum \infty \frac{10 n}{4 n + 1}

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