What is the sum from n=0 to infinity of 8/n-8/(n+1) ?

The sum from n=0 to infinity of 8/n-8/(n+1) is 8

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sum from n=0 to infinity of 8/n-8/(n+1)

n = 0 \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

5^{^{'}}

\int x e^{\frac{1}{2} x^{2}} d x

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

lo g_{9} (ln (x))

\frac{d}{d x} (1 - 4 x)

What is the sum from n=0 to infinity of 8/n-8/(n+1) ?
The sum from n=0 to infinity of 8/n-8/(n+1) is 8