What is the sum from n=0 to infinity of n/(2n^2+3) ?

The sum from n=0 to infinity of n/(2n^2+3) is diverges

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sum from n=0 to infinity of n/(2n^2+3)

n = 0 \sum \infty \frac{n}{2 n ^{2} + 3}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{2 n ^{2} + 3}

Apply Series Integral Test:

diverges

= diverges

n = 3 \sum \infty \frac{1}{n ^{100}}

n = 0 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

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

j = 1 \sum \infty 3^{- 4 j}

n = 1 \sum \infty ln (\frac{10}{n})

What is the sum from n=0 to infinity of n/(2n^2+3) ?
The sum from n=0 to infinity of n/(2n^2+3) is diverges