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

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

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

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

8

Solution steps

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

Apply Telescoping Series Test:

8

= 8

\int \frac{6 x - 11}{( x - 1 ) ^{2}} d x

\frac{6}{( 6 - x ^{6} ) ^{\frac{3}{2}}}

\int \frac{x ^{3}}{( x ^{2} + 4 ) ^{\frac{3}{2}}} d x

\frac{6}{( s + 3 ) ^{2}}

\int (1 - 3 x)^{\frac{1}{2}} d x

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