I'm trying to prove this by induction: $$\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}<\frac{3}{2}, \quad n\geq1. $$ I got stuck in the inductive step: $$\frac{3}{2} +\frac{1}{(n+1)^3}\,\ldots$$ Thank you for your help!

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    $\begingroup$ There's not going to be an inductive step that takes you directly from $S_n < 3/2$ to $S_{n+1}=S_{n}+1/(n+1)^3 < 3/2$, since the latter doesn't follow from the former. You need to come up with something tighter as your inductive hypothesis. For instance, can you show that $S_n < 3/2 - F_n$ implies $S_{n+1} < S_{n} + 1/(n+1)^3 < 3/2 - F_{n+1}$ for some positive error function $F_n$ of your choice? $\endgroup$ – mjqxxxx Oct 5 '14 at 20:41
  • $\begingroup$ @mjqxxxx Ok I see, then the point is to find such $F_n$... Thank you! $\endgroup$ – N. Abel Oct 5 '14 at 20:44
  • $\begingroup$ @mjqxxxx: I'd add this as the answer, because the difficult step is realising that by using a tighter hypothesis, you need to prove something stronger. but also have something stronger to prove it with. Finding $F_n$ should be easy. $\endgroup$ – gnasher729 Oct 9 '14 at 22:15

You can use for $n\geq 2$, $$\dfrac{1}{n^3} \leq \dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$$

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    $\begingroup$ Brilliant :) (+1) $\endgroup$ – AlexR Oct 5 '14 at 20:46
  • $\begingroup$ @Liu Gang Please can you elaborate more on this? thanks. $\endgroup$ – N. Abel Oct 5 '14 at 20:48
  • $\begingroup$ @N.Abel By this inequality we get your LHS is smaller than $1 + \sum_{k=2}^n(\frac{1}{k} - \frac{1}{k+1}) = \frac{3}{2} - \frac{1}{n+1} $ $\endgroup$ – Petite Etincelle Oct 5 '14 at 20:50
  • $\begingroup$ @LiuGang Thank you very much! $\endgroup$ – N. Abel Oct 5 '14 at 20:50

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