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

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

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

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

3

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Telescoping Series Test:

1

= 3 \cdot 1

Simplify

= 3

x \to 0 lim (\frac{( 1 + x ) ^{2}}{x})

\int \frac{( ln ( x ) ) ^{2} + 1}{ln ( x )} d x

y^{^{'}} = \frac{e ^{x - y}}{e ^{x} + 1}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + k y = 0

y^{^{'}} + 3 x e^{y} = 0

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