`sum_(n=1)^oo sin((2n-1)pi/2)` Determine the convergence or divergence of the series.
`sin ((2n-1) pi/2)` is a function that alternates from 1 to -1, and is in fact equivalent to `(-1)^(n+1)`.
Thus this series is 1-1+1-1+1-1...
Because `lim_(n->oo) (-1)^(n+1) != 0` since the terms alternate between -1 and 1, then by the n'th term Test, the series diverges.
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