# 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?

• Yes it is right. – Aditya De Saha Apr 18 '17 at 6:23
• yes it is right ! – Hptunjy Prjkeizg Apr 18 '17 at 10:34

You can also use Cauchy condensation: it converges iff $\sum_{n=2}^{\infty} \frac{1}{2^n (n-1)}$ converges, but that converges by comparison with $\frac{1}{2^n}$.
You can also use the ratio test: the ratio is $\frac{2^n-1}{2^{n+1}-1} \to \frac{1}{2}$.
You can also use comparison with the convergent $\sum_{n=1}^{\infty} 2 \times \frac{1}{2^n}$.
You can also use the root test: $$\lim_{n \to \infty} \left(\frac{1}{2^n-1}\right)^{1/n} = \frac{1}{2}$$ though that limit is a bit tedious.
$$\begin{array}{rcccl} \dfrac1{2^n} &\le& \dfrac{1}{2^n - 1} &\le& \dfrac1{2^{n-1}} \\ \displaystyle \sum_{n=1}^\infty \dfrac1{2^n} &\le& \displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n - 1} &\le& \displaystyle \sum_{n=1}^\infty \dfrac1{2^{n-1}} \\ 1 &\le& \displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n - 1} &\le& 2 \end{array}$$