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

Question

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.

Answer 1

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.

Answer 2

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: $$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right) < \frac{3}{2}\sqrt{e} <\frac{5}{2}.$$

Answer 3

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*}

Answer 4

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.

Answer 5

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

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