What is the sum from n=1 to infinity of 4/(n^2+3n+2) ?

The sum from n=1 to infinity of 4/(n^2+3n+2) is 2

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sum from n=1 to infinity of 4/(n^2+3n+2)

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

2

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

\frac{1}{2}

= 4 \cdot \frac{1}{2}

Simplify

4 \cdot \frac{1}{2} : 2

= 2

\frac{d}{d x} (2 e^{\frac{s i n ( x )}{3}} + 1)

sin (\frac{π}{2} - x)

x \to - 1 - lim (x^{2})

x \to 3 - lim (\frac{x ^{2}}{x - 3})

\int 6 x^{\frac{2}{3}} d x

What is the sum from n=1 to infinity of 4/(n^2+3n+2) ?
The sum from n=1 to infinity of 4/(n^2+3n+2) is 2