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

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

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

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

diverges

Solution steps

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

Apply Series Comparison Test:

diverges

= 2 diverges

= diverges

k = 2 \sum \infty \frac{k}{ln ( k )}

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

n = 8 \sum \infty \frac{1}{n ^{2} + 64}

n = 1 \sum \infty \frac{1}{3 ( 2 ^{n} )}

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

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