Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations and worked out examples convince me that this inequality indeed holds true. I have tried to solve this problem by induction. Clearly, for $N=1$ we have,
$$\mathcal{P}(1) = \left(\frac{2}{3}\right)^{\tfrac{3}{4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{4}} \approx 0.8774 < 1.$$
Now assume the inequality holds for $N$, then for $N+1$ we have,
\begin{align}
\mathcal{P}(N+1) &=\left(\frac{2N+2}{2N+3}\right)^{\tfrac{2N+3}{2N+4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{2N+4}} \\
 &= \left(\left(\frac{2N}{2N+1}\right)\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)\right)^{\left(\frac{2N+1}{2N+2}\right)\left(\frac{2N+2}{2N+1}\cdot\frac{2N+3}{2N+4}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\frac{2N+2}{2N+4}}\\[1em]
&= \left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)\left(1 + \frac{1}{2(N+1/2)(N+2)}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\left(1 - \frac{1}{N+2}\right)}\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \small\left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}}\cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \mathcal{P}(N) \cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \cdot\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}
\end{align}
Now from here we know that the first three terms are all smaller than 1 ($\mathcal{P}(N) < 1$ by induction hypothesis). However the last term is larger than one. For the proof by induction to work out, we need that this last term cancels against,
$$\left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}}.$$ But I do not see how it does. Any help is greatly appreciated.
 A: We have
$$
(\mathcal{P}(N))^{2N + 2}  = \left( {\frac{{2N}}{{2N + 1}}} \right)^{2N + 1} 2 = \frac{1}{{\left( {1 + \frac{1}{{2N}}} \right)^{2N} }}\frac{2}{{1 + \frac{1}{{2N}}}} < \frac{4}{9} \cdot 2 < 1,
$$
since $
{\left( {1 + \frac{1}{n}} \right)^n }
$ is increasing monotonically to $e$.
A: You can use Young's inequality to prove it
$$(\frac{2N}{2N+1})^{\frac{2N+1}{2N+2}}(\frac{2}{1})^{\frac{1}{2N+2}}\leq \frac{1}{\frac{2N+2}{2N+1}}((\frac{2N}{2N+1})^{\frac{2N+1}{2N+2}})^{\frac{2N+2}{2N+1}}+\frac{1}{2N+2}((\frac{2}{1})^{\frac{1}{2N+2}})^{2N+2}=1$$
the two sides are equal iff $((\frac{2N}{2N+1})^{\frac{2N+1}{2N+2}})^{\frac{2N+2}{2N+1}}=((\frac{2}{1})^{\frac{1}{2N+2}})^{2N+2}$, which is impossible
A: If you could show that it is an increasing function of real positive $N$ and its limit is $1$ then that would be enough.
Showing the limit is $1$ as $N \to \infty$ looks easy enough: each component heads towards $1$
while the derivative seems to be $\mathcal P'(x)=\frac{{{\left( \frac{x}{2x+1}\right) }^{\frac{2x+1}{2x+2}}}\left( x\mathrm{log}\left( \frac{x}{2x+1}\right) +x+1\right) }{x{{\left( x+1\right) }^{2}}}$ which seems to be positive for positive $x$, so $\mathcal P(x)$ must be strictly less than $1$ for all positive real $x$ and thus $\mathcal P(N)$ must be strictly less than $1$ for all positive integer $N$
A: It is equivalent to show that $\log \mathcal{P}(N) < 0$ for $N \ge 1$. That is,
$$
\begin{align*}
\frac{2N+1}{2N+2}\log\left(\frac{2N}{2N + 1}\right) + \frac{\log(2)}{2N+2} &< 0 \\
\frac{2N+1}{2N+2}\log\left(1-\frac{1}{2N + 1}\right) + \frac{\log(2)}{2N+2} &< 0 
\end{align*}
$$
Since for $0 < x  < 1$ we have $\log(1-x) < -x - \frac{x^2}{2}$,
$$
\log\left(1-\frac{1}{2N + 1}\right) < -\frac{1}{2N+1} - \frac{1}{2(2N+1)^2} < -\frac{1}{2N+1} - \frac{1}{(2N+1)^2}
$$
Thus
$$
\begin{align*}
\frac{2N+1}{2N+2}\log\left(1-\frac{1}{2N + 1}\right) + \frac{\log(2)}{2N+2} &< \frac{2N+1}{2N+2}\left(-\frac{1}{2N+1} - \frac{1}{(2N+1)^2}\right) + \frac{\log(2)}{2N+2}\\
&= \frac{(2N+1)(\log(2)-1) - 1}{(2N+1)(2N+2)}  < 0
\end{align*}
$$
since $\log(2) - 1 < 0$.
A: Written in a more compact form
$$P_n=2 \left(\frac{n}{2 n+1}\right)^{1-\frac{1}{2 (n+1)}}$$
Take logarithms and compose Taylor series
$$\log(P_n)=\frac{\log (2)-1}{2 n}+\frac{3-4 \log (2)}{8
   n^2}+O\left(\frac{1}{n^3}\right)$$
Continue with Taylor
$$P_n=e^{\log(P_n)}=1-\frac{1-\log (2)}{2 n}+\frac{4+(\log (2)-6) \log (2)}{8
   n^2}+O\left(\frac{1}{n^3}\right)$$
Even for $n=1$, the above truncated expansion is not bad. In fact, the relative error is less than $0.01$% as soon as $n \geq 5$.
A: We present a proof that utilises a special case of the well known Bernoulli's Inequality.
Lemma:
$\forall \ x \geq 0, (1+x)^n \geq 1+nx, n\in \mathbb{N}.$
(This can easily be proven by applying the Binomial Theorem)
Now,
\begin{align}
&\left(\dfrac{2N}{2N+1}\right)^\dfrac{2N+1}{2N+2} \cdot \left(\dfrac{2}{1}\right)^\dfrac{1}{2N+2}  < 1 \\
& \iff \left(\dfrac{2N}{2N+1}\right)^{2N+1} \cdot 2  < 1 \\
& \iff \left(\dfrac{2N+1}{2N}\right)^{2N+1} \cdot \dfrac{1}{2}  > 1 \\
& \iff \left(1+\dfrac{1}{2N}\right)^{2N+1} > 2 ​
\end{align}
The last inequality follows immediately from our lemma - since $\dfrac{1}{2N} > 0 \ \forall \ N \in \mathbb{N}$, $\left(1+\dfrac{1}{2N}\right)^{2N+1} \geq 1+\dfrac{2N+1}{2N} >2.$
