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  • 5 years ago

Convergent or divergent series? sigma (n=1 to infinity) sin (npi/2)/n

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  1. anonymous
    • 5 years ago
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    This is a convergent series. It's an alternating series. If you expand on a few terms, you see: S = 1 - (1/3) + (1/5) - (1/7) + (1/9) - ... So the series has the form \[\sum_{n=0}^{\infty}(-1)^na_n=\]\[\sum_{n=0}^{\infty}(-1)^n{\frac{1}{2n+1}}\]where all the a[n] are non-negative. If the limit of the sequence a[n] equals 0 as n approaches infinity, and the sequence an is monotone decreasing, then the series converges by the Alternating Series Test (otherwise known as the Leibniz Criterion for Alternating Series). The sum is, incidentally, \[\frac{\pi}{4}\]Hope this helps.

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