What is the sum from n=1 to infinity of 6/(4n^2+4n) ?

The sum from n=1 to infinity of 6/(4n^2+4n) is 3/2

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sum from n=1 to infinity of 6/(4n^2+4n)

n = 1 \sum \infty \frac{6}{4 n ^{2} + 4 n}

\frac{3}{2}

Decimal

1.5

Solution steps

n = 1 \sum \infty \frac{6}{4 n ^{2} + 4 n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} + 4 n}

Apply Telescoping Series Test:

\frac{1}{4}

= 6 \cdot \frac{1}{4}

Simplify

6 \cdot \frac{1}{4} : \frac{3}{2}

= \frac{3}{2}

What is the sum from n=1 to infinity of 6/(4n^2+4n) ?
The sum from n=1 to infinity of 6/(4n^2+4n) is 3/2