Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$
This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)
 A: Denote $p(n):=(1+\frac{1}{1^3})(1+\frac{1}{2^3})(1+\frac{1}{3^3})...(1+\frac{1}{n^3})$. 

Claim:
$$p(n)\leq3-\frac2{n^2},\,\forall n\geq2.$$

For $n=2$, we have $\frac94\leq3-\frac12$.
Then suppose $p(n)\leq3-\frac2{n^2}$ for some $n$. We see
$$\eqalign{
p(n+1)&=p(n)(1+\frac1{(n+1)^3})\cr
&\leq3-\frac{2}{n^2}+\frac3{(n+1)^3}-\frac{2}{n^2(n+1)^3}\cr
&=3+\frac{3n^2-2(n^3+3n^2+3n+1)-2}{n^2(n+1)^3}\cr
&=3-\frac{2n^3+3n^2+6n+4}{n^2(n+1)^3}\cr
&=3-\frac{2n^3+2n^2+(n^2+6n+4)}{n^2(n+1)^3}\cr
&\leq3-\frac{2n^2(n+1)}{n^2(n+1)^3}\cr
&\leq3-\frac2{(n+1)^2}}.$$

As pointed out by @saulspatz, one can prove that $p(n)\leq3-\frac1n,\forall n\geq1$ by the same method.

Hope this helps.
A: We have that
$$\prod_{k=1}^\infty \left(1+\frac{1}{k^3}\right)<3\iff \sum_{k=1}^\infty  \log\left(1+\frac{1}{k^3}\right)<\log 3$$
and since $\forall x>0\, \log(1+x)<x$
$$\sum_{k=1}^\infty  \log\left(1+\frac{1}{k^3}\right)=\log 2+\sum_{k=2}^\infty  \log\left(1+\frac{1}{k^3}\right)<\log 2+\sum_{k=2}^\infty  \frac{1}{k^3}<\log 3$$
A: Taking seriously the suggestions of Dr. Sonnhard Graubner and J.G., one can indeed prove that, for $x\in\mathbb{C}$ and $p\in\mathbb{N}_+$,
$$
\prod_{n=1}^\infty \left(1+\frac{x^p}{n^p}\right) =
\prod_{j=1}^p \frac{1}{\Gamma(1+\omega_p^j x)} ,
$$
where $\{-\omega_p^j\}_{j=1}^p$ are the $p$-th roots of $-1$.
In fact, using that $\sum_{j=1}^p \omega_p^j = 0$ and this identity (see also this), one has that
$$
\prod_{n=1}^N \left(1+\frac{x^p}{n^p}\right) =
\prod_{j=1}^p \prod_{n=1}^N \left(1+\frac{\omega_p^j x}{n}\right) =
\prod_{j=1}^p e^{-\omega_p^j x \gamma}
    \prod_{n=1}^N \left(1+\frac{\omega_p^j x}{n}\right) e^{-\frac{\omega_p^j x}{n}}
$$
converges, as $N\to\infty$, to
$$
\prod_{j=1}^p \frac{1}{\omega_p^j x \Gamma(\omega_p^j x)} =
\prod_{j=1}^p \frac{1}{\Gamma(1+\omega_p^j x)}.
$$
Specializing for $p=3$ and $x=1$ and using the formula for the absolute value here, we have
$$
\prod_{n=1}^\infty \left(1+\frac{1}{n^p}\right) =
\frac1{\Gamma(1+1)\Gamma(1-\frac12-\frac{\sqrt3 i}2)\Gamma(1-\frac12+\frac{\sqrt3 i}2)} = \frac1{|\Gamma(\frac12+\frac{\sqrt3 i}2)|} = \frac{\cosh\left(\frac{\sqrt3 }2\pi\right)}{\pi} .
$$
A: With $AM-GM$
\begin{align}
1.(1+\frac{1}{2^3})(1+\frac{1}{3^3})\cdots(1+\frac{1}{n^3}) 
&\leq\left(\dfrac1n(n+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots+\frac{1}{n^3})\right)^n \\
&\leq\left(\dfrac1n(n+\sum_{n=1}^\infty\frac{1}{n^3}-1)\right)^n \\
&\leq\left(\dfrac1n(n+\zeta(3)-1)\right)^n \\
&\leq\left(1+\dfrac{\zeta(3)-1}{n}\right)^n \\
&< e^{\zeta(3)-1}\\
&<\frac32
\end{align}
Thanks to Winther.
A: By induction we can prove the stronger 
$$\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$$
indeed
1. base cases: by inspection the inequality is satisfied for for
    $n=1,2, 3$
2. induction step: 


*

*assume true that (Ind. Hyp.): $\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$

*we want to prove that: $\prod_{k=1}^{n+1}\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n+1}\right)$
then we have
$$\prod_{k=1}^{n+1}\left(1+\frac{1}{k^3}\right)=\prod_{k=1}^n\left(1+\frac{1}{k^3}\right) \cdot \left(1+\frac{1}{(n+1)^3}\right)<$$
$$\stackrel{Ind. Hyp.}<3\left(1-\frac{1}{n}\right)\left(1+\frac{1}{(n+1)^3}\right)\stackrel{?}<3\left(1-\frac{1}{n+1}\right)$$
thus we need to show that
$$3\left(1-\frac{1}{n}\right)\left(1+\frac{1}{(n+1)^3}\right)\stackrel{?}<3\left(1-\frac{1}{n+1}\right)$$
which is true indeed
$$1+\frac{1}{{n+1}^3}-\frac1n-\frac{1}{n(n+1)^3}\stackrel{?}<1-\frac{1}{n+1}$$
$$n-(n+1)^3-1\stackrel{?}<-n(n+1)^2$$
$$n-n^3-3n^2-3n-1-1\stackrel{?}<-n^3-2n^2-n$$
$$n^2+n+2\stackrel{?}>0$$
A: Claim:

For all positive integers $n$, we have
$$\prod_{k=1}^n \left(1+\frac{1}{k^3}\right) < e$$
or equivalently
$$\sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right) < 1$$
Proof:

It suffices to show that
$$\sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right) < 1-\frac{1}{(n+1)^2}\tag{*}$$
holds for all positive integers $n$.

To prove $(*)$, proceed by induction on $n$.

By direct evaluation, $(*)$ holds for the base case $n=1$.

Suppose $(*)$ holds for some positive integer $n$.
\begin{align*}
\text{Then}\;\;&\sum_{k=1}^{n+1} \ln\left(1+\frac{1}{k^3}\right)\\[4pt]
=\;&
\left(
\sum_{k=1}^n \ln\left(1+\frac{1}{k^3}\right)
\right)
+
\ln\left(1+\frac{1}{(n+1)^3}\right)
\\[4pt]
<\;&
\left( 1-\frac{1}{(n+1)^2}\right)
+
\ln\left(1+\frac{1}{(n+1)^3}\right)
&&\text{[by the inductive hypothesis]}
\\[4pt]
<\;&
\left( 1-\frac{1}{(n+1)^2}\right)
+
\frac{1}{(n+1)^3}
&&\text{[since $\ln(1+x) < x$, for all $x > 0$]}
\\[4pt]
=\;&\left(1-\frac{1}{(n+2)^2}\right)-\frac{n^2+n-1}{(n+1)^3(n+2)^2}
\\[4pt]
<\;&1-\frac{1}{(n+2)^2}
&&\text{[since $n^2+n-1 > 0$]}
\\[4pt]
\end{align*}
which completes the induction, and thus proves the claim.
A: Less Than $\boldsymbol{3}$
The inequality
$$
1+\frac1{n^3}\lt\frac{1+\frac1{2(n-1)^2}}{1+\frac1{2n^2}}\tag1
$$
can be verified by cross-multiplying and then multiplying both sides by $2n^5(n-1)^2$; that is,
$$
2n^7-4n^6+3n^5\underbrace{-3n^3+3n^2-2n+1}_\text{$-(3n^2+2)(n-1)-1\lt0$ for $n\ge1$}\lt2n^7-4n^6+3n^5\tag2
$$
Therefore, employing a telescoping product,
$$
\begin{align}
\prod_{n=1}^\infty\left(1+\frac1{n^3}\right)
&\lt2\prod_{n=2}^\infty\frac{1+\frac1{2(n-1)^2}}{1+\frac1{2n^2}}\\
&=2\cdot\frac32\\[9pt]
&=3\tag3
\end{align}
$$

Actual Value
$$
\begin{align}
\lim_{n\to\infty}\prod_{k=1}^n\frac{k^3+1}{k^3}
&=\lim_{n\to\infty}\frac{\Gamma(n+2)\,\Gamma\!\left(n+\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(n+\frac12-i\frac{\sqrt3}2\right)}{\Gamma(2)\,\Gamma\!\left(\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(\frac12-i\frac{\sqrt3}2\right)\Gamma(n+1)^3}\tag4\\
&=\frac1{\Gamma\!\left(\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(\frac12-i\frac{\sqrt3}2\right)}\\
&\times\lim_{n\to\infty}\frac{\Gamma(n+2)\,\Gamma\!\left(n+\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(n+\frac12-i\frac{\sqrt3}2\right)}{\Gamma(n+1)^3}\tag5\\
&=\frac{\sin\left(\frac\pi2+i\frac{\pi\sqrt3}2\right)}{\pi}\times1\tag6\\[6pt]
&=\frac{\cosh\left(\frac{\pi\sqrt3}2\right)}{\pi}\tag7
\end{align}
$$
Explanation:
$(4)$: $\prod\limits_{k=1}^n(k+x)=\frac{\Gamma(n+1+x)}{\Gamma(1+x)}$ and $k^3+1=(k+1)\left(k-\frac12+i\frac{\sqrt3}2\right)\left(k-\frac12-i\frac{\sqrt3}2\right)$
$(5)$: pull out the constant factor using $\Gamma(2)=1$
$(6)$: apply Euler's Reflection Formula $\Gamma(x)\,\Gamma(1-x)=\frac\pi{\sin(\pi x)}$
$\phantom{(6)\text{:}}$ and Gautschi's Inequality, which implies $\lim\limits_{n\to\infty}\frac{\Gamma(n+x)}{\Gamma(n)\,n^x}=1$
$(7)$: $\cos(ix)=\cosh(x)$
