<s>[INST] Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$. Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$.
I'm not sure, but I suppose that $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < \sum_{n=0}^{\infty} \frac{1}{2^n} = 2$$
I don't know how can I prove it exactly, but we have:
$$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} = 1+ \frac{1}{3} + \frac{1}{7} + \frac{1}{15}+... \\
\sum_{n=0}^{\infty} \frac{1}{2^n} = 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+...$$
Because
$$1 \le 1 \\ \frac{1}{3} < \frac{1}{2} \\ \frac{1}{7} < \frac{1}{4} \\ \frac{1}{15} < \frac{1}{8} \\ \mbox{and so on}$$
we have
$$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} = 1+ \frac{1}{3} + \frac{1}{7} + \frac{1}{15}+... <  1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+... = \sum_{n=0}^{\infty} \frac{1}{2^n}$$
Could you tell me, does it work? Maybe exist more formal proof?
 [/INST] Notice that 
$$\sum_{n=0}^\infty \frac{1}{2^n}=\sum_{n=1}^\infty \frac{1}{2^{n-1}}=2$$
and the result follows since
$$\frac{1}{2^{n-1}}\geq \frac{1}{2^{n}-1}\iff 2^{n-1}\geq1,\; \forall n\geq1\quad\text{which's true}$$
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