<s>[INST] Do the following series converge or diverge? Justify. $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$$
Can I do this using limit comparison test?
for $n \in [1, \infty), a_n = \frac{1}{2^n - 1} \geq 0$
for $n \in [1,\infty), b_n = \frac{1}{2^n} \geq 0$
Since $$\lim_{n\to\infty} \frac{2^n}{2^n-1} = 1 \in (0,\infty)$$
$\sum_{n=1}^{\infty} \frac{1}{2^n}$. This is a geometric series with $r = \frac{1}{2} < 1$. Therefore by the geometric series this $\sum b_n$ converges. So does $\sum a_n$ by the limit comparison test. 
Is this right? 
 [/INST] By setting boundaries:
$$\begin{array}{rcccl}
\dfrac1{2^n} &\le& \dfrac{1}{2^n - 1} &\le& \dfrac1{2^{n-1}} \\
\displaystyle \sum_{n=1}^\infty \dfrac1{2^n} &\le& \displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n - 1} &\le& \displaystyle \sum_{n=1}^\infty \dfrac1{2^{n-1}} \\
1 &\le& \displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n - 1} &\le& 2
\end{array}$$
</s>