What is the sum from n=1 to infinity of (5n)/(n^2+4) ?

The sum from n=1 to infinity of (5n)/(n^2+4) is diverges

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sum from n=1 to infinity of (5n)/(n^2+4)

n = 1 \sum \infty \frac{5 n}{n ^{2} + 4}

diverges

Solution steps

n = 1 \sum \infty \frac{5 n}{n ^{2} + 4}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{n}{n ^{2} + 4}

Apply Series Comparison Test:

diverges

= 5 diverges

= diverges

y = x^{2} - 3 x - 5, (2, - 7)

\frac{d}{d x} (\frac{6000}{x})

\int sin^{6} (x) cos^{3} (x) d x

x \to 0 lim (x^{7 x})

\frac{3}{s ^{2} + 16}

What is the sum from n=1 to infinity of (5n)/(n^2+4) ?
The sum from n=1 to infinity of (5n)/(n^2+4) is diverges