What is the sum from n=8 to infinity of 1/n-1/(n+3) ?

The sum from n=8 to infinity of 1/n-1/(n+3) is converges

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sum from n=8 to infinity of 1/n-1/(n+3)

n = 8 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

converges

Solution steps

n = 8 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

Expand

\frac{1}{n} - \frac{1}{n + 3} : \frac{3}{n ^{2} + 3 n}

= n = 8 \sum \infty \frac{3}{n ^{2} + 3 n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 8 \sum \infty \frac{1}{n ^{2} + 3 n}

Apply Series Comparison Test:

converges

= 3 converges

= converges

n = 3 \sum \infty (1 + n)^{- 1}

n = 1 \sum \infty e^{- \frac{n}{3}}

n = 1 \sum \infty \frac{8 ^{n}}{n + 1}

n = 0 \sum \infty \frac{1}{( n + 2 ) n !}

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

What is the sum from n=8 to infinity of 1/n-1/(n+3) ?
The sum from n=8 to infinity of 1/n-1/(n+3) is converges