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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

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

n = 1 \sum \infty \frac{e ^{n}}{n ^{8}}

n = 0 \sum \infty (1 + \frac{2}{n})^{n}

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

n = 1 \sum \infty \frac{3 n - 2}{n}

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