What is the sum from n=1 to infinity of 8/(6n^2+6n) ?

The sum from n=1 to infinity of 8/(6n^2+6n) is 4/3

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sum from n=1 to infinity of 8/(6n^2+6n)

n = 1 \sum \infty \frac{8}{6 n ^{2} + 6 n}

\frac{4}{3}

Solution steps

n = 1 \sum \infty \frac{8}{6 n ^{2} + 6 n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{6 n ^{2} + 6 n}

Apply Telescoping Series Test:

\frac{1}{6}

= 8 \cdot \frac{1}{6}

Simplify

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

= \frac{4}{3}

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

n = 2 \sum \infty \frac{7}{3 ^{n}}

n = 2 \sum \infty 2^{- 4 n}

n = 0 \sum \infty \frac{n}{6 + n ^{4}}

n = 1 \sum \infty n^{5} e^{n^{6}}

What is the sum from n=1 to infinity of 8/(6n^2+6n) ?
The sum from n=1 to infinity of 8/(6n^2+6n) is 4/3