I'm reading Tom Apostol's calculus and I'm a bit stuck on proving an inequality which is part of the proof of theorem 1.15 (the integral of $x^p$).

$$ \sum_{k=1}^{n-1}k^p < \frac{n^{p+1}}{p+1} $$

Now the author says this can be proven by induction, but I have been trying for a few days without success! I'm sure it's pretty straightforward but for some reason I cannot get to it.

I would really be grateful if someone could help me finding the way out of this!

Thanks Fabrizio

  • $\begingroup$ It would be really great if you share your work and the manner you approached it! $\endgroup$ – kishlaya Jan 16 '17 at 11:34
  • $\begingroup$ $p$ is positive? $\endgroup$ – joseabp91 Jan 16 '17 at 11:44
  • $\begingroup$ yes $p \geq 1$. $\endgroup$ – user3093226 Jan 16 '17 at 11:59

Hint. In the inductive step you have to show that $$\frac{n^{p+1}}{p+1}+n^p\leq \frac{(n+1)^{p+1}}{p+1}$$ that is $$(p+1)n^p\leq (n+1)^{p+1}-n^{p+1},$$ or $$\frac{p+1}{n}\leq \left(1+\frac{1}{n}\right)^{p+1}-1,$$ then use Bernoulli's inequality.

  • $\begingroup$ Isn't the mean value theorem a lot to assume for someone studying introductory calculus? $\endgroup$ – shredalert Jan 16 '17 at 11:48
  • $\begingroup$ Hi Robert, I think shredalert is right. At this stage I'm not supposed to know what MVT is. I got exactly at the point you said above, but I'm not sure how to show the inductive step! Thanks $\endgroup$ – user3093226 Jan 16 '17 at 11:59
  • $\begingroup$ Fine. You can use also Bernoulli's inequality. See my edited answer.. $\endgroup$ – Robert Z Jan 16 '17 at 12:03
  • $\begingroup$ Excellent, thanks! $\endgroup$ – user3093226 Jan 16 '17 at 12:11

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