What is the sum from n=7 to infinity of 1/(n^2-5n+6) ?

The sum from n=7 to infinity of 1/(n^2-5n+6) is 1/4

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sum from n=7 to infinity of 1/(n^2-5n+6)

n = 7 \sum \infty \frac{1}{n ^{2} - 5 n + 6}

\frac{1}{4}

Solution steps

n = 7 \sum \infty \frac{1}{n ^{2} - 5 n + 6}

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

n = 1 \sum \infty \frac{5}{n ^{9}}

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

n = 1 \sum \infty \frac{1}{( 6 n ) ^{6}}

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

n = 1 \sum \infty (- 1)^{n} + 1

What is the sum from n=7 to infinity of 1/(n^2-5n+6) ?
The sum from n=7 to infinity of 1/(n^2-5n+6) is 1/4