Quote:
Originally Posted by Ether
Is there a shorter way?
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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: S
k=1OO (k/(k+3))
2k converges
Therefore, S
k=1OO (k/(k+3))
k2 converges by the Direct Comparison Test.