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

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

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

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

converges

Solution steps

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

Simplify

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

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

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

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

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

x = 0 \sum \infty \frac{x ^{x}}{x !}

n = 0 \sum \infty \frac{n}{n ^{2} + 3}

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