I was testing for convergence of a series:(adsbygoogle = window.adsbygoogle || []).push({});

∑[itex]\frac{1}{n^2 -1}[/itex] from n=3 to infinity

I used the integral test, substituting n as 2sin(u)

so here's the question:

when using the trig substitution, I realized the upperbound, infinity, would fit inside the sine.

Is it still possible to make the substitution? Or is there a restriction when this happens?

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# Trig substitution into integrals

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