Prove that $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)<3$ for all positive integers $n$

This problem is copied from Math Olympiad Treasures by Titu Andreescu and Bogdan Enescu.They start by stating that induction wouldn't directly work here since the right hand side stays constant while the left increases.They get rid of this problem by strengthening the hypothesis.$$\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)\le3-\dfrac{1}{n}$$ and then proceed by induction.

The problem is that I can't find a motivation for the above change.I mean,we could have subtracted a lot of things from the RHS but what should nudge us to try $\dfrac{1}{n}$?The rest of the proof is quite standard,but I can't see how I am supposed to have thought of it.Is it just experience?Or is it a standard technique?A little guidance and motivation will be appreciated.

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    $\begingroup$ It won't be Olympiad problem if you can easily find the trick. :p $\endgroup$
    – user175968
    Commented Nov 6, 2014 at 12:50
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    $\begingroup$ The "subtract something and prove a stronger inequality" is standard. Then you say you want to prove $p_n < 3 - a_n$, and check what (preferably simple) $a_n$ would work for the induction step. That won't always work (at least not well), but often does. $\endgroup$ Commented Nov 6, 2014 at 12:52
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    $\begingroup$ Usually you can have insights by looking at the first terms. You take $n=1,2,3,4$... and try to see if you can deduce something out of it. $\endgroup$
    – Martigan
    Commented Nov 6, 2014 at 12:52
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    $\begingroup$ I hope that the authors also provided the solution $$\prod_{k=1}^n \left( 1 + \frac{1}{k^3}\right) = 2 \prod_{k=2}^n \frac{k^3+1}{k^3} < 2\prod_{k=2}^n \frac{k^3+1}{k^3-1} < 2\prod_{k=2}^\infty \frac{k^3+1}{k^3-1} = 3$$ with the telescoping product. $\endgroup$ Commented Nov 6, 2014 at 12:58
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    $\begingroup$ How does it telescope exactly? I don't really see it since all terms are different.. $\endgroup$
    – mvggz
    Commented Nov 6, 2014 at 13:10

6 Answers 6


As mentioned in the comments, the trick to subtract something (or multiply with something $< 1$) and prove a stronger inequality than required is standard.

So we want to prove

$$p_n \leqslant 3- a_n$$

for a preferably simple $a_n > 0$ by induction. For the induction start, we here need $a_1 \leqslant 1$. For the induction step to work, we need

$$\left( 1 + \frac{1}{(n+1)^3}\right)(3-a_n) \leqslant 3 - a_{n+1}.$$

Multiplying out and cancelling the $3$, we get

$$-a_n + \frac{3}{(n+1)^3} - \frac{a_n}{(n+1)^3} \leqslant -a_{n+1}.$$

Throwing away the $\frac{a_n}{(n+1)^3}$ term to simplify the calculations, we see that

$$\frac{3}{(n+1)^3} \leqslant a_n - a_{n+1}$$

is sufficient. With the ansatz $a_n = \frac{c}{n^k}$, we have

$$a_n - a_{n+1} = c\frac{(n+1)^k-n^k}{n^k(n+1)^k} \approx c\frac{k}{n(n+1)^k},$$

so here we need $k \leqslant 2$. But $a_1 \leqslant 1$ requires $c\leqslant 1$, and hence $k = 2$ doesn't work. So we try $k = 1$ and find $a_n = \frac{1}{n}$ works.

  • $\begingroup$ Thank you very very much for this answer.Do you mind if I use this answer(with all credits to you) somewhere else?It would help some other people on Brilliant.org if I could.And thanks again for this answer. $\endgroup$
    – rah4927
    Commented Nov 7, 2014 at 7:51
  • $\begingroup$ No, I don't mind at all. Go ahead and share. $\endgroup$ Commented Nov 7, 2014 at 12:52

Since: $$ 1+\frac{1}{k^3}=\left(1+\frac{1}{k}\right)\left(1-\frac{1}{k}+\frac{1}{k^2}\right) = \left(1-\frac{1}{k^2}\right)\left(1+\frac{1}{k(k-1)}\right) $$ and: $$ \prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)=\frac{1}{2} $$ we have: $$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)=\prod_{k=2}^{+\infty}\left(1+\frac{1}{k(k-1)}\right)=\prod_{k=1}^{+\infty}\left(1+\frac{1}{k(k+1)}\right),$$ but since $1+x < e^x$ and $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ it follows that: $$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)<\exp\sum_{k=1}^{+\infty}\frac{1}{k(k+1)}=e<3.$$ With the same trick we can also prove the stronger: $$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right) < \frac{3}{2}\sqrt{e} <\frac{5}{2}.$$

  • $\begingroup$ How does this even remotely answer the question? $\endgroup$ Commented Nov 6, 2014 at 14:16
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    $\begingroup$ It is just an interesting technique (a telescoping one) to prove a stronger upper bound. $\endgroup$ Commented Nov 6, 2014 at 14:20

If we need to prove the strong inequality and proving the weak inequalities for the simplest to imagine values is easy...

\begin{align*} \left(3-\frac1n\right)\left(1+\frac1{(n+1)^3}\right) &= 3+\frac3{(n+1)^3}-\frac1n-\frac1{n(n+1)^3}\\ &=3+\frac{3n-(n+1)^3-1}{n(n+1)^3}\\ &=3+\frac{-n^3-3n^2-2}{n(n+1)^3}\\ &=3+\frac{-1}{n+1} \frac{n^3+3n^2+2}{n^3+2n^2+n}\\ &\leq3-\frac1{n+1} \end{align*}

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    $\begingroup$ And how was $\frac 1n$ chosen? $\endgroup$
    – abiessu
    Commented Nov 6, 2014 at 13:16
  • $\begingroup$ @abiessu We have product less than $3-\frac1n$, for $n$, then we compute the estimation of product for $n+1$. $\endgroup$ Commented Nov 6, 2014 at 13:19
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    $\begingroup$ @PrzemysławScherwentke,I am afraid you misread the question.The question was asking for some motivation regarding the choice of $\frac{1}{n}$. $\endgroup$
    – rah4927
    Commented Nov 6, 2014 at 13:20
  • $\begingroup$ @rah4927 Oh! I thought that showing, how easy is such approximation, is a good motivation. :-( $\endgroup$ Commented Nov 6, 2014 at 13:23

Just for the fun of it, I tried to use this (I've answered about it not long ago :) ) :

$ [x_1*...*x_n]^{\frac{1}{n}} \leq \frac{x_1+...+x_n}{n} $

Now set: $ x_k = 1 + \frac{1}{k^3} $; $C_n = \sum_{k=2}^{n} \frac{1}{k^3} $

You get : $x_1*...*x_n = 2*(1+\frac{1}{2^3})...(1 +\frac{1}{n^3}) \leq 2*(\frac{x_2+...+x_n}{n-1})^{n-1} = 2*(\frac{n-1+C_n}{n-1})^{n-1} $

=> $ 2*(1+\frac{1}{2^3})...(1 +\frac{1}{n^3}) \leq 2*(1+\frac{1}{n-1}*C_n)^{n-1} $

Now $C_n \rightarrow C $, with $C_n \leq C $; and $ (1+\frac{C_n}{n-1})^{n-1} \leq e^{C_n} \leq e^C$

Hence you get:$ 2*(1+\frac{1}{2^3})...(1 +\frac{1}{n^3}) \leq 2*e^C = e^{ln(2) + C}$

Now we want to see if : $ln(2) + C \leq ln(3) $ Numerically this is true since $ln(\frac{3}{2})$ ~0,4~$2*C$. Not very difficult to prove.

It's not really what was asked but I thought it was interesting.


Just to give a different approach (similar to Jack D'Aurizio's), note that, for $k\gt1$, we have

$$\ln\left(1+{1\over k^3}\right)\lt{1\over k^3}\lt{1\over k^3-k}={1\over2}\left({1\over k-1}-{2\over k}+{1\over k+1}\right)$$

It follows that

$$\begin{align} \ln\left(\left(1+{1\over1^3}\right)\left(1+{1\over2^3}\right)\left(1+{1\over3^3}\right)\cdots\right) &=\ln2+\sum_{k=2}^\infty\ln\left(1+{1\over k^3}\right)\\ &\lt\ln2+{1\over2}\sum_{k=2}^\infty\left({1\over k-1}-{2\over k}+{1\over k+1}\right)\\ &=\ln2+{1\over2}\left({1\over2-1}-{2\over2}+{1\over3-1} \right)\quad\text{(after telescoping)}\\ &=\ln2+{1\over4}\\ &={1\over4}\ln(16e)\\ &\lt{1\over4}\ln81\\ &=\ln3 \end{align}$$


When n is smaller than 3, it is easy to compute it.

When n is greater or equal to 3:

$$ \begin{aligned} \ln \left[(1+\frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3}) \right] &= \ln 2+ \sum_{i=2}^{n}\ln(1+\frac{1}{i^3}) \\ &<\ln 2 +\sum_{i=2}^{n}\frac{1}{i^3} \\ &<\ln 2+\frac{1}{8}+\int_2^n \frac{1}{x^3} dx \\ &=\ln 2+\frac{1}{8}+\left[ -\frac{1}{2x^2} \right]_2^n \\ &=\ln 2+\frac{1}{4}-\frac{1}{2n^2} \\ &<\ln 2 + \frac{1}{4}<1<\ln 3 \end{aligned} $$


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