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

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

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

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

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

n = 1 \sum \infty \frac{1}{n ^{1.2}}

n = 2 \sum \infty (- 0.15)^{n}

n = 0 \sum \infty \frac{20}{4 ^{n}}

n = 1 \sum \infty (\frac{π}{3})^{n}

k = 1 \sum \infty \frac{1}{k + 1}

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