show that $$\sum\_{k=2}^{n}\frac{1}{3^k-1}<\dfrac{1}{5}\tag1$$

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> I try to use this well known: if $a>b>0,c>0$,then we have
>  $$\dfrac{b}{a}<\dfrac{b+c}{a+c}$$
>  $$\dfrac{1}{3^k-1}<\dfrac{1+1}{3^k-1+1}=\dfrac{2}{3^k}$$
>  so we have
>  $$\sum\_{k=2}^{n}\dfrac{1}{3^k-1}<\sum\_{k=2}^{n}\dfrac{2}{3^k}=\dfrac{1}{3}$$but this is big than $\dfrac{1}{5}$,so how to prove inequality (1)
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