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I am stuck with the this question:

Prove by induction that $2^n > n^3$, for all $n \ge 10$

I got this far:

Base: For $P(10)$: $$ 2^n > n^3 \\ 2^{10} > 10^3 \\ 1024 > 1000 $$ so, $P(10)$ is true

Inductive steps: Need to show $P(k+1)$ is true, assuming $P(k)$ is true.

For $P(k+1)$: $$ 2^{k+1} > (k+1)^3 \\ 2 \times 2^k > k^3 + 3k^2 + 3k + 1 \\ 2^k + 2^k > k^3 + 3k^2 + 3k + 1 $$

We assume $2^k > k^3$, so we only need to show $2^k > 3k^2 + 3k + 1$ ? But I am stuck here.

Any idea/hints? Thanks a lot for the help.

Michael Hardy
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user350954
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1 Answers1

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Since $k\ge10$, $(1+1/k)^3=((1+1/k)^k)^{3/k}\le (e)^{1/3}\lt2$.
Hence, if $2^k\gt k^3$, then $2^{k+1}\gt2k^3\gt k^3(1+1/k)^3=(k+1)^3$.
So the proof is complete.

awllower
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  • I did not find that there was a duplicate... – awllower Jul 09 '13 at 02:42
  • thanks for the answer but I haven't learned e, so I cannot follow your thought – user350954 Jul 09 '13 at 02:44
  • The last solution in the linked problem has a similar character, since it uses ratios. – André Nicolas Jul 09 '13 at 02:45
  • @AndréNicolas Yes, I guess that answer is essentially the same as mine, while not availing of the definition of $e$. – awllower Jul 09 '13 at 02:47
  • @user350954 Maybe you can try to prove that $\lim_{k\to\infty}(1+1/k)^k$ exists, and that this limit is less than $8$! I think this is contained in every calculus text. If necessary, I can do it, I think. :) – awllower Jul 09 '13 at 02:50