<s>[INST] Prove this inequality $\frac{n}{3^n-2^n}<\left(\frac{2}{5}\right)^{n-1}$ Let $n\ge 3$. How to prove show that
$$\dfrac{n}{3^n-2^n}<\left(\dfrac{2}{5}\right)^{n-1}\tag1$$
or 
$$3^n\cdot 2^{n-1}-2^{2n-1}-5^{n-1}\cdot n>0,n\ge 3\tag2$$
It seem right:see wolfram
when I do this problem found it:How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$
if this inequality (1) has prove it,then
$$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<1+\dfrac{2}{5}+\sum_{k=3}^{+\infty}\dfrac{2^{k-1}}{5^{k-1}}=\dfrac{5}{3}$$
Question: How to prove inequality (1) or (2)
 [/INST] $(A)\enspace$ Induction
$n=3$ : $\enspace$ That’s clear (or see $(B)$) .
Be $\enspace\displaystyle \left(\frac{6}{5}\right)^n - \left(\frac{4}{5}\right)^n - \frac{2n}{5} > 0 \enspace$ (equivalent to $(1)$)$\enspace$ for $\enspace n\geq 4\,$ .
It follows:  
$\displaystyle \left(\frac{6}{5}\right)^{n+1} - \left(\frac{4}{5}\right)^{n+1} - \frac{2(n+1)}{5} = $
$\displaystyle =\left(\frac{6}{5}\right)^n - \left(\frac{4}{5}\right)^n - \frac{2n}{5} + \frac{1}{5}\left(\left(\frac{6}{5}\right)^n + \left(\frac{4}{5}\right)^n -2\right) $ 
$\displaystyle > \frac{1}{5}\left(\left(\frac{6}{5}\right)^4 -2\right) = \frac{1}{5}(2.0736 - 2) > 0 $ 

$(B)\enspace$ Derivation
$x=3$ : $\enspace\displaystyle \frac{3}{19}=\frac{x}{3^x-2^x}|_{x=3}<\left(\frac{2}{5}\right)|_{x=3}=0.16$   
$x=4$ : $\enspace\displaystyle \frac{4}{65}=\frac{x}{3^x-2^x}|_{x=4}<\left(\frac{2}{5}\right)|_{x=4}=0.064$   
$x\in\mathbb{R}$ with $x\geq 5$ , $(1)$ multiplicated by $(3^x-2^x)\frac{2}{5}$ :
$\enspace\enspace\displaystyle \frac{2x}{5} < \left(\frac{6}{5} \right)^x - \left(\frac{4}{5}\right)^5 \leq \left(\frac{6}{5} \right)^x - \left(\frac{4}{5}\right)^x $
The right side is clear, the left side we proof by derivation.
$x=5$ : $\enspace\displaystyle 2 = \frac{2x}{5}|_{x=5} < \left(\frac{6}{5}\right)^x|_{x=5} -  \left(\frac{4}{5}\right)^5 = 2.16064 $
It’s  
$\displaystyle \frac{2}{5} = \frac{d}{dx} \frac{2x}{5} < \frac{d}{dx} (\left(\frac{6}{5}\right)^x - \left(\frac{4}{5}\right)^5 ) = \left(\frac{6}{5}\right)^x \ln \frac{6}{5} $
because of  
$\displaystyle \left(\frac{6}{5}\right)^x \geq \left(\frac{6}{5}\right)^5 = 2.48832 >  \frac{2}{5} / \ln \frac{6}{5} \approx 2.1939… \enspace$ .  
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