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

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

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

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

diverges

Solution steps

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

Apply Series Comparison Test:

diverges

= diverges

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

n = 0 \sum \infty (1 + \frac{2}{5 n})^{2}

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

n = 1 \sum \infty \frac{n}{n + 200}

n = 0 \sum \infty \frac{1}{1000 n + 1}

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