<s>[INST] How prove this $\sum_{k=2}^{n}\frac{1}{3^k-1}<\frac{1}{5}$ show that  $$\sum_{k=2}^{n}\frac{1}{3^k-1}<\dfrac{1}{5}\tag1$$

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)

 [/INST] Note that for $n\ge2$, 
$$\begin{align}\sum_{k=2}^n\frac{1}{3^k-1}&=\frac{1}{8}\end{align}+\sum_{k=3}^n\frac{1}{3^k-1}<\frac{1}{8}+\sum_{k=3}^n\frac{1}{3^k-3}=\frac{1}{8}+\frac{1}{3}\sum_{k=2}^{n-1}\frac{1}{3^k-1}$$
so you can use induction.
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