# Prove that $(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3$ [duplicate]

I have tried to use induction, but after I assume that P(n) is true, I can't go further to prove that P(n+1) is true as well. I also have tried to find an intermediate inequality, but I can't figure out which inequality I should start from.

Something that seemed to be useful was taking P(n) and multiplying it by $$(1+\frac{1}{(n+1)^3})$$, therefore I have come to this

$$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3 | \times(1+\frac{1}{(n+1)^3})$$

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

but, as anyone could imagine, I came to contradiction because I've tried to prove that $$3(1+\frac{1}{(n+1)^3})<3$$ which is false.

Any help it would be useful.

• Dec 4, 2020 at 20:24

Using the fact $$1+x\le e^x$$ for all real $$x,$$ we have $$\left(1+ \frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\cdots\left(1+\frac{1}{n^3}\right)\le \dfrac{9}{4}\exp\left(\sum_{k=3}^{n}\dfrac{1}{k^3}\right).$$ Now use the fact that $$\sum_{k=1}^{\infty}\dfrac{1}{k^3}\lt\dfrac{\pi^2}{7}.$$