Let $n\in\Bbb N$ and $x>0$. Prove Prove that $n\ln\left (\dfrac{1+nx}{nx}\right )<1/x$.
Try: Trying to play with inequality. $$\begin{align*}\ln\left (\dfrac{1+nx}{nx}\right )<1/(nx)&\Rightarrow \dfrac{1+nx}{nx}<e^{1/(nx)}\\ &\Rightarrow 1+nx<nx\cdot e^{1/(nx)} \end{align*}$$ Let $u=nx$, then $1+u<ue^{1/u}$. Any help, after that?
You want to prove $1+u<ue^{\frac{1}{u}}$. To show this it is enough to show that $1+\frac{1}{u}<e^{\frac{1}{u}}$. Now by the Taylor series expansion of $e^x$ we have
$e^{\frac{1}{u}}=1+\frac{1}{u}+\frac{1}{2!u^2}+\frac{1}{3!u^3}+\cdots>1+\frac{1}{u}$, for $u>0$.
I have yet to encounter an inequality about the exponential that cannot be reduced to $$e^x\ge 1+x\qquad\text{for all }x\in\Bbb R\text{ with equality iff }x=0. $$
Simply , as , $u=nx \gt 0 $ ,$e^{1/u}=1+\frac{1}{1! u}+\frac{1}{2! u^2}+\cdot\cdot\cdot $ is greater than $1+\frac{1}{u}$.
And, hence your conclusion.
