How prove this $\sum_{k=2}^{n}\frac{1}{3^k-1}<\frac{1}{5}$

Question

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)

Answer 1

Since $3^k > 6$ for $k \geq 2$, $$\frac{1}{3^k-1}<\frac{1}{3^k-\frac{3^k}{6}}=\frac{6}{5}\cdot\frac{1}{3^k}$$ so $$\sum_{k=2}^{n}\dfrac{1}{3^k-1}<\frac{6}{5}\sum_{k=2}^{n}\dfrac{1}{3^k}<\frac{6}{5}\sum_{k=2}^\infty\dfrac{1}{3^k}=\dfrac{1}{5}$$ as required.

Answer 2

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.

Answer 3

$$\sum_{k\geq 2}\frac{1}{3^k-1}=-\frac{1}{2}+\sum_{k\geq 1}\sum_{m\geq 1}\frac{1}{3^{km}}=-\frac{1}{2}+\sum_{n\geq 1}\frac{d(n)}{3^n}\leq -\frac{1}{2}+\sum_{n=1}^{5}\frac{d(n)}{3^n}+\sum_{n\geq 6}\frac{n}{3^n}$$ leads to the sharper inequality $$ \mathcal{T}=\sum_{k\geq 2}\frac{1}{3^k-1}\leq \frac{61}{324}.$$

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We may actually devise an acceleration method for the series

$$\mathcal{S}=\sum_{n\geq 1}\frac{1}{3^n-1}=\sum_{n\geq 1}\frac{1}{3^n}+\sum_{n\geq 1}\frac{1}{3^n(3^n-1)}=\frac{1}{2}+\mathcal{T}$$ since $$ \frac{1}{9}\mathcal{T} = \sum_{n\geq 1}\frac{1}{3^{n+1}(3^{n+1}-3)},\qquad \mathcal{T}-\frac{1}{6}=\sum_{n\geq 1}\frac{1}{3^{n+1}(3^{n+1}-1)} $$ are pretty close, namely $$ \frac{1}{6}-\frac{8}{9}\mathcal{T}=\sum_{n\geq 1}\frac{2}{3^{n+1}(3^{n+1}-1)(3^{n+1}-3)}\geq \frac{1}{216}$$ (leading to $\mathcal{T}\leq\frac{35}{192}$) but $$ \frac{1}{27}\left(\frac{1}{6}-\frac{8}{9}\mathcal{T}\right)-\left(\frac{1}{6}-\frac{8}{9}\mathcal{T}-\frac{1}{216}\right)=16\sum_{n\geq 1}\frac{1}{3^{n+2}(3^{n+2}-1)(3^{n+2}-3)(3^{n+2}-9)}\geq \frac{1}{18954} $$ leads to $\mathcal{T}\geq \frac{11821}{64896}$, hence $\mathcal{T}=\color{green}{0.182}\ldots$.