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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ^{0.6}}

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

n = 4 \sum \infty n e^{- n^{2}}

n = 0 \sum \infty - \frac{1}{n}

n = 0 \sum \infty cos (\frac{nπ}{2})

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