What is the sum from n=1 to infinity of 3/(n^2+n) ?

The sum from n=1 to infinity of 3/(n^2+n) is 3

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sum from n=1 to infinity of 3/(n^2+n)

n = 1 \sum \infty \frac{3}{n ^{2} + n}

3

Solution steps

n = 1 \sum \infty \frac{3}{n ^{2} + n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + n}

Apply Telescoping Series Test:

1

= 3 \cdot 1

Simplify

= 3

\int \frac{4 x ^{2} + 6}{x ^{3} + 3 x} d x

f (x) = 7 x - 1

y^{^{''}} + 4 y^{^{'}} + 6 y = 0

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

y = ln (\frac{e ^{x}}{e ^{x} - 1})

What is the sum from n=1 to infinity of 3/(n^2+n) ?
The sum from n=1 to infinity of 3/(n^2+n) is 3