What is the sum from n=1 to infinity of 2/(n^2+n) ?

The sum from n=1 to infinity of 2/(n^2+n) is 2

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sum from n=1 to infinity of 2/(n^2+n)

n = 1 \sum \infty \frac{2}{n ^{2} + n}

2

Solution steps

n = 1 \sum \infty \frac{2}{n ^{2} + n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + n}

Apply Telescoping Series Test:

1

= 2 \cdot 1

Simplify

= 2

x \to 0 lim (\frac{( x - 1 ) ^{3} + 1}{h ( x )})

x \to + 1 lim (x sin (\frac{1}{x}))

\frac{\partial}{\partial y} (\frac{y}{y + x})

\int_{- 2}^{0} x + 2 d x

f (x) = 2 e^{2 x}

What is the sum from n=1 to infinity of 2/(n^2+n) ?
The sum from n=1 to infinity of 2/(n^2+n) is 2