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Prove n^2/2^n converges and find its limit?
I understand that it converges because 2^n grows a lot faster than n^2 and I think the limit is zero but how do I go about proving it?
2 Answers
- ?Lv 78 years agoFavorite Answer
To prove it, show that for any small e>0(your limit), you can find a value within e of 0, that is n^2/2n<|e|
- alwbsokLv 78 years ago
L'Hopital's rule (twice) should do the trick. Recall d/dx 2^x = ln(2) * 2^x, and the limit is in an infinity/infinity indeterminate form, so the rule applies. An equivalent limit is:
2n / (ln(2) * 2^n)
Using it again:
2 / (ln(2)^2 * 2^n)
which tends to 0.