What is the sum from n=1 to infinity of 4/(n(n+1)) ?

The sum from n=1 to infinity of 4/(n(n+1)) is 4

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sum from n=1 to infinity of 4/(n(n+1))

n = 1 \sum \infty \frac{4}{n ( n + 1 )}

4

Solution steps

n = 1 \sum \infty \frac{4}{n ( n + 1 )}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

1

= 4 \cdot 1

Simplify

= 4

\frac{d}{d x} (ln (x) + \frac{1}{ln ( x )})

x \to 0 lim (1 + 2 x)

\frac{\partial}{\partial x} (x e^{x^{2} y^{2}})

x \to 1 lim (b + \frac{1}{x - 1})

What is the sum from n=1 to infinity of 4/(n(n+1)) ?
The sum from n=1 to infinity of 4/(n(n+1)) is 4