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

The sum from n=1 to infinity of 3/n-3/(n+2) is 9/2

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

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

\frac{9}{2}

Solution steps

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

Simplify

\frac{3}{n} - \frac{3}{n + 2} : \frac{6}{n ( n + 2 )}

= n = 1 \sum \infty \frac{6}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n ( n + 2 )}

Take the partial fraction of

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

= 6 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

6 \cdot \frac{3}{4} : \frac{9}{2}

= \frac{9}{2}

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

n = 0 \sum \infty \frac{1}{5 + n}

k = 3 \sum \infty \frac{1}{4 k ^{2} - 1}

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

n = 2 \sum \infty \frac{n !}{( 2 n ) !}

What is the sum from n=1 to infinity of 3/n-3/(n+2) ?
The sum from n=1 to infinity of 3/n-3/(n+2) is 9/2