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

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

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

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

converges

Solution steps

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

Simplify

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

n = 0 \sum \infty \frac{n}{11 n + 1}

m = 2 \sum \infty \frac{3}{4 ^{m}}

n = 1 \sum \infty \frac{n ^{2}}{5 + 2 n}

n = 1 \sum \infty cos (\frac{3}{n ^{2}})

k = 4 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

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