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

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

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

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

\frac{1}{3}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{3}

= \frac{1}{3}

n = 2 \sum \infty \frac{1}{4 n ^{3}}

n = 2 \sum \infty \frac{2 ^{n}}{n !}

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

n = 1 \sum \infty 0. 5^{n - 1}

n = 1 \sum \infty π^{\frac{1}{n}}

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