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.

  • $\begingroup$ 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

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 $$\frac1{2^3}+\ldots\frac1{n^3}\le\frac58-\frac1n\!\!\!$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.