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I need to prove that: $$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} < \frac{3}{2}\sqrt[3]{n^2}$$

The problem is that I can't find a way to separate that "$1$" on the right side ($\frac{3}{2}\sqrt[3]{(n+1)^2}$) to get a sum on the let AND right side. I thought about Bernoulli's inequality but it doesn't seem to help much either.

Edit: I've got some bomb idea, I will post it as a comment

theboyboy
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3 Answers3

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You seek to verify the inequality $\frac{3}{2} \sqrt[3]{n^2} + \frac{1}{\sqrt[3]{n+1}} \overset{?}{<} \frac{3}{2} \sqrt[3]{(n+1)^2}.$ Cubing both sides yields $$n^2 + 3 (2/3) n^{4/3}(n+1)^{-1/3} + 3(2/3)^2 n^{2/3}(n+1)^{-2/3} + (2/3)^3 (n+1)^{-1/3} \overset{?}{<} (n+1)^2$$ $$3 (2/3) n^{4/3}(n+1)^{-1/3} + 3(2/3)^2 n^{2/3}(n+1)^{-2/3} + (2/3)^3 (n+1)^{-1/3}\overset{?}{<} 2n + 1$$

Note that the first term is $< n$, the second term is $<4/3$, and the last term is $<(2/3)^3 2^{-1/3}< 1/3$ for $n \ge 2$. So the left-hand side is bounded by $ < n+\frac{5}{3} < 2n+1$ for $n \ge 2$.

angryavian
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  • I have edited my question - it should be fine now. What do you mean by all those terms? I thought it has to ben done with induction but apparently I was wrong. – theboyboy Nov 27 '20 at 22:28
  • @theboyboy The first inequality in my answer is what you need after you set up the induction. – angryavian Nov 27 '20 at 22:30
  • Ok and what about those terms? For example: what is "$<n$"? – theboyboy Nov 27 '20 at 22:31
  • @theboyboy What I mean is the first term on the left-hand side ($3(2/3)n^{4/3}(n+1)^{-1/3}$) is smaller than $n$. – angryavian Nov 27 '20 at 22:32
  • Ohhhh, now I get it – theboyboy Nov 27 '20 at 22:34
  • That's great, I need to remember to check terms separatelly when I can not compose my induction proof. – theboyboy Nov 27 '20 at 22:35
  • This answer is completely wrong (I cubed the inequality incorrectly, and also the first term is approximately $2n$, not $<n$). I don't think crudely bounding terms like this will be tight enough; last inequality in my answer is true, but very tight. I'm flagging a moderator to delete this answer since I cannot delete it myself. – angryavian May 05 '21 at 14:32
  • @theboyboy Can you un-accept this answer? – angryavian May 05 '21 at 14:33
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Since $k^{-1/3}=\int_{k-1}^kk^{-1/3}dx<\int_{k-1}^kx^{-1/3}dx$,$$\sum_{k=1}^nk^{-1/3}<\sum_{k=1}^n\int_{k-1}^kx^{-1/3}dx=\int_0^nx^{-1/3}dx=\tfrac32n^{2/3}.$$

J.G.
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  • That's a 2nd mistake in one question. Can the proof be done without integrals? I mean it should be possible, but I was not able to do that. – theboyboy Nov 27 '20 at 22:17
  • @theboyboy Hint for an inductive proof: asymptotically, $\left(n+a\right)^{p}\in n^{p}\left(1+\frac{ap}{n}+\frac{a^{2}p\left(p-1\right)}{2n^{2}}+O\left(\frac{1}{n^{3}}\right)\right)$, so$$\frac{3}{2}\left(\left(n+1\right)^{2/3}-n^{2/3}\right)-\left(n+1\right)^{-1/3}\in\frac{1}{6}n^{-4/3}\left(1-\frac{4}{3n}+O\left(\frac{1}{n^{2}}\right)\right).$$ – J.G. Nov 27 '20 at 22:28
  • Thank you, but I am afraid I don't understand. I like the solution provided by angryavian. He uses separate terms because it is apparently not possible to solve that by induction. – theboyboy Nov 27 '20 at 22:39
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So I got stuck at that point with my induction:

$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \frac{3}{2}\sqrt[3]{(n+1)^2}$$

Then, trying to separate elements on the right hand side I got:

$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \frac{3}{2}\sqrt[3]{n^2(1 + \frac{2}{n} + \frac{1}{n^2})}$$ $$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \frac{3}{2}\sqrt[3]{n^2}\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}$$ And now I see that the growth of values on the right hand side associated with changes of that ($\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}$) part is the smallest when that ($\frac{1}{\sqrt[3]{n}}$) part is the smallest and it is allways bigger than 1. So I know that: $$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}<\frac{3}{2}\sqrt[3]{n^2}\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}$$

Now I see that:

  • $ \frac{1}{\sqrt[3]{n+1}} $ is falling for all $ n \in \mathbb{N}$
  • $\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}$ is growing for all $ n \in \mathbb{N}$

So I just check both valuse for n = 1 and I get accordingly:

  • $ \frac{1}{\sqrt[3]{2}} < 1$
  • $\sqrt[3]{4} > 1$

So, that is it. To make sure I can calculate:

  • $\lim_{x \to +\infty} \frac{1}{\sqrt[3]{n+1}} = 0$
  • $\lim_{x \to +\infty}\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})} = \infty $
theboyboy
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