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

The sum from n=1 to infinity of 1/n-1/(n+3) is 11/6

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

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

\frac{11}{6}

Solution steps

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

Expand telescoping series:

\frac{11}{6}

= \frac{11}{6}

y = 0, y = 2 x^{2}, y = 12 x - 4 x^{2}, [0, 2]

((6 x + 1) y^{2}) \frac{d y}{d x} + 3 x^{2} + 2 y^{3} = 0

x \to - 4 lim (0)

\int_{4}^{\infty} e^{3 x} d x

\int tan^{5} (x) sec^{10} (x) d x

What is the sum from n=1 to infinity of 1/n-1/(n+3) ?
The sum from n=1 to infinity of 1/n-1/(n+3) is 11/6