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

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

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

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

diverges

Solution steps

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

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} + 2}

Apply Series Integral Test:

diverges

= 5 diverges

= diverges

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\frac{d y}{d x} - e^{x + y} = 0

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x^{2} - 1 x y + x \frac{d y}{d x} = 0

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