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

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

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

n = 4 \sum \infty \frac{1}{n ^{2} - 5 n + 6}

1

Solution steps

n = 4 \sum \infty \frac{1}{n ^{2} - 5 n + 6}

Apply Telescoping Series Test:

1

= 1

x \to - 4 lim (- \frac{4}{x + 4})

e^{x}, 4

\frac{d}{d x} (\frac{x}{3 x - 4 ln ( x )})

\frac{\partial}{\partial x} ((- 3 x y) 0)

\frac{d}{d t} (V (t))

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