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

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

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

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

3

Solution steps

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

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} + 2 n}

Take the partial fraction of

\frac{1}{n ^{2} + 2 n} : \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

= 4 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

4 \cdot \frac{3}{4} : 3

= 3

n = 1 \sum \infty \frac{1}{( n ( n + 1 ) )}

n = 0 \sum \infty (\frac{1}{2})^{2}

k = 1 \sum \infty \frac{1}{k ^{2} + k}

n = 0 \sum \infty \frac{cos ( nπ )}{5}

n = 2 \sum \infty 7 e^{- 3 n}

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