What is the sum from n=1 to infinity of n/(2n^2+3) ?

The sum from n=1 to infinity of n/(2n^2+3) is diverges

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sum from n=1 to infinity of n/(2n^2+3)

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{n + 6}

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

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

n = 1 \sum \infty \frac{1}{n ^{1.43}}

n = 1 \sum \infty \frac{9}{7 n ^{2} + 7 n}

What is the sum from n=1 to infinity of n/(2n^2+3) ?
The sum from n=1 to infinity of n/(2n^2+3) is diverges