What is the sum from n=6 to infinity of 1/n-1/(n+2) ?

The sum from n=6 to infinity of 1/n-1/(n+2) is converges

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sum from n=6 to infinity of 1/n-1/(n+2)

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

converges

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

n = 0 \sum \infty c n

n = 1 \sum \infty - 5 n

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

n = 1 \sum \infty \frac{4 ^{n}}{n ^{5}}

n = 1 \sum \infty 7^{n}

What is the sum from n=6 to infinity of 1/n-1/(n+2) ?
The sum from n=6 to infinity of 1/n-1/(n+2) is converges