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

The sum from n=2 to infinity of 6/(n(n+2)) is converges

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

n = 2 \sum \infty \frac{6}{n ( n + 2 )}

converges

Solution steps

n = 2 \sum \infty \frac{6}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 6 \cdot n = 2 \sum \infty \frac{1}{n ( n + 2 )}

Apply Series Comparison Test:

converges

= 6 converges

= converges

n = 2 \sum \infty (- 0.45)^{n}

n = 1 \sum \infty 7 - 4 (0.8)^{n}

n = 0 \sum \infty \frac{1}{9 - n ^{2}}

n = 1 \sum \infty 3 (\frac{1}{7})^{n + 1}

k = 3 \sum \infty \frac{k - 1}{2 k + 1}

What is the sum from n=2 to infinity of 6/(n(n+2)) ?
The sum from n=2 to infinity of 6/(n(n+2)) is converges