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

The sum from n=2 to infinity of 4/(n(n+2)) is converges

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

n = 2 \sum \infty \frac{4}{n ( n + 2 )}

converges

Solution steps

n = 2 \sum \infty \frac{4}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 4 \cdot n = 2 \sum \infty \frac{1}{n ( n + 2 )}

Apply Series Comparison Test:

converges

= 4 converges

= converges

k = 1 \sum \infty \frac{k}{e ^{k}}

n = 0 \sum \infty (1 + \frac{3}{n})^{n}

n = 1 \sum \infty \frac{1}{5 ^{n} + 1}

n = 1 \sum \infty \frac{ln ( n ^{2} )}{n}

n = 1 \sum \infty \frac{5}{n ^{2} + 6}

What is the sum from n=2 to infinity of 4/(n(n+2)) ?
The sum from n=2 to infinity of 4/(n(n+2)) is converges