Thread: Math Quiz 7
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Unread 27-03-2015, 21:12
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Re: Math Quiz 7

Quote:
Originally Posted by Ether View Post
Is there a shorter way?
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.
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Last edited by GeeTwo : 27-03-2015 at 21:16.
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