What is the sum from n=1 to infinity of 8/(n+2)-8/n ?

The sum from n=1 to infinity of 8/(n+2)-8/n is -12

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sum from n=1 to infinity of 8/(n+2)-8/n

n = 1 \sum \infty \frac{8}{n + 2} - \frac{8}{n}

- 12

Solution steps

n = 1 \sum \infty \frac{8}{n + 2} - \frac{8}{n}

Expand

\frac{8}{n + 2} - \frac{8}{n} : - \frac{16}{n ^{2} + 2 n}

= n = 1 \sum \infty - \frac{16}{n ^{2} + 2 n}

Apply the constant multiplication rule:

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

= - 16 \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 )}

= - 16 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

- 16 \cdot \frac{3}{4} : - 12

= - 12

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

n = 1 \sum \infty 6^{- n}

k = 2 \sum \infty (- 0.45)^{k}

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

n = 0 \sum \infty ln (\frac{n}{3 n + 1})

What is the sum from n=1 to infinity of 8/(n+2)-8/n ?
The sum from n=1 to infinity of 8/(n+2)-8/n is -12