What is the sum from n=0 to infinity of 1/(n+2)-1/n ?

The sum from n=0 to infinity of 1/(n+2)-1/n is -3/2

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sum from n=0 to infinity of 1/(n+2)-1/n

n = 0 \sum \infty \frac{1}{n + 2} - \frac{1}{n}

- \frac{3}{2}

Solution steps

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

Expand

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

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

Apply the constant multiplication rule:

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

= - 2 \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 )}

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

Expand telescoping series:

\frac{3}{4}

Simplify

- 2 \cdot \frac{3}{4} : - \frac{3}{2}

= - \frac{3}{2}

n = 0 \sum \infty \frac{e ^{n}}{n ^{4}}

n = 1 \sum \infty \frac{1}{5 n + 3}

n = - \infty \sum \infty n

n = 1 \sum \infty \frac{3 !}{3 ^{n}}

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

What is the sum from n=0 to infinity of 1/(n+2)-1/n ?
The sum from n=0 to infinity of 1/(n+2)-1/n is -3/2