What is the sum from n=0 to infinity of (2n)/(n^2+4) ?

The sum from n=0 to infinity of (2n)/(n^2+4) is diverges

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sum from n=0 to infinity of (2n)/(n^2+4)

n = 0 \sum \infty \frac{2 n}{n ^{2} + 4}

diverges

Solution steps

n = 0 \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 = 0 \sum \infty \frac{n}{n ^{2} + 4}

Apply Series Integral Test:

diverges

= 2 diverges

= diverges

n = 0 \sum \infty \frac{1}{( n ln ( n ) )}

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

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

n = 0 \sum \infty \frac{8}{( - 3 ) ^{n}}

n = 1 \sum \infty n \cdot 0. 5^{n}

What is the sum from n=0 to infinity of (2n)/(n^2+4) ?
The sum from n=0 to infinity of (2n)/(n^2+4) is diverges