What is the sum from n=1 to infinity of 9/n-9/(n+1) ?

The sum from n=1 to infinity of 9/n-9/(n+1) is 9

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sum from n=1 to infinity of 9/n-9/(n+1)

n = 1 \sum \infty \frac{9}{n} - \frac{9}{n + 1}

9

Solution steps

n = 1 \sum \infty \frac{9}{n} - \frac{9}{n + 1}

Apply Telescoping Series Test:

9

= 9

n = 1 \sum \infty \frac{6}{3 n ^{2} + 2}

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

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

n = 0 \sum \infty n^{2} e^{- n^{3}}

n = 1 \sum \infty \frac{5}{n !}

What is the sum from n=1 to infinity of 9/n-9/(n+1) ?
The sum from n=1 to infinity of 9/n-9/(n+1) is 9