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Re: Math Quiz 7
For all k>2, 0 < k/(k+3) < 1
For all k>2, k[sup]2 > 2k Therefore, for all k>2, (k/(k+3))k2 < (k/(k+3))2k. Already proven: Sk=1OO (k/(k+3))2k converges Therefore, Sk=1OO (k/(k+3))k2 converges by the Direct Comparison Test. Last edited by GeeTwo : 27-03-2015 at 21:16. |
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