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Use strong induction to prove that

$$\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}\leq\frac{5}{8}-\frac{1}{n}$$ $$n\geq2$$

I'm not sure how to go about this. I used base cases n=2, and n=3 but I'm having trouble with the actual induction part.

I said this as my Induction Hypothesis: Suppose the claim is true for n=2, 3, ... ,k. We must show it holds for n=k+1.

Not sure where to go from there :T.

Thanks in advance for help.

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  • $\begingroup$ Not that it is helpful for this question but the infinite sum $\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}+\cdots$ is $1$ less than Apéry's constant so less than $0.20205690316$ $\endgroup$ – Henry Jul 22 '18 at 17:21
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Assuming it holds for $n-1$ it suffices to show that $$\frac1{n^3}\le\frac1{n-1}-\frac1n=\frac1{n(n-1)}$$ which is obviously true.

You should see that adding \begin{align*}\frac1{2^3}+\ldots\frac1{(n-1)^3}&\le\frac58-\frac1{n-1}\\ \frac1{n^3}&\le\frac1{n-1}-\frac1n\end{align*} yields $$\frac1{2^3}+\ldots\frac1{n^3}\le\frac58-\frac1n\!\!\!$$

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  • $\begingroup$ Could you explain this in more detail? Sorry, but this is a bit confusing to me. How you get to the conclusion, to be specific. $\endgroup$ – Chris Jang Apr 15 '14 at 20:06
  • $\begingroup$ @ChrisJang What conclusion? $\endgroup$ – user2345215 Apr 15 '14 at 20:08
  • $\begingroup$ The part after "yields" . I get lost after that. Do you use substitution, or...? $\endgroup$ – Chris Jang Apr 15 '14 at 20:14
  • $\begingroup$ @ChrisJang That's easy. Just compute the sum both inequalities above. (left sides and right sides separately) $\endgroup$ – user2345215 Apr 15 '14 at 20:16
  • $\begingroup$ OOOOHHH Okay, I understand now; you add the two inequalities. Thank you very much! :) $\endgroup$ – Chris Jang Apr 15 '14 at 20:27

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