What is the sum from n=1 to infinity of 7/(n(n+1)) ?

The sum from n=1 to infinity of 7/(n(n+1)) is 7

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sum from n=1 to infinity of 7/(n(n+1))

n = 1 \sum \infty \frac{7}{n ( n + 1 )}

7

Solution steps

n = 1 \sum \infty \frac{7}{n ( n + 1 )}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

1

= 7 \cdot 1

Simplify

= 7

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

f (x) = 3 (x + 1)

\frac{d y}{d x} - y = x e^{x}

x \to 3 lim (3 x^{2} - 7 x + 2)

x \to 5 + lim (- \frac{x}{x - 5})

What is the sum from n=1 to infinity of 7/(n(n+1)) ?
The sum from n=1 to infinity of 7/(n(n+1)) is 7