Quote:
Originally Posted by GeeTwo
How about this one: - [as shown earlier] (k/(k+3))2k (the terms, not the series sum) converges to e-6 ~ .00248 ~ 1/403.
- This means that for every k sufficiently large, (k/(k+3))2k < 1/4 (actually I believe this is true for all of them)
- As k/(k+3) is positive, (k/(k+3))k < 1/2 for sufficiently large k.
- Therefore, for every k sufficiently large, (k/(k+3))k2 < 1/2k
- We know Sk=1OO 1/2k, converges at 1.0.
- By the comparison test, Sk=1OO (k/(k+3))[sup]k2[/sup converges.
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Unless I'm missing something, I see no flaw there. Nice.