Does $$\sum_{n=1}^{\infty} \dfrac{1}{2^n -1}$$ converge or diverge?
The tests I know so far are: by defintion, geometric, divergence, integral.
The only appealing method right now is the integral test, but the it looks hard to integrate. Any suggestions? I don't see how I can rewrite this in an equivalent form.
For all $n\ge 1$ we have $2^{n-1}+1\le 2^{n-1}+2^{n-1}=2^n$, thus $$2^{n-1}\le 2^n-1\iff \frac{1}{2^n-1}\le\frac{1}{2^{n-1}}$$ for all $n\ge 1$.
Then we have $$\sum_{n=1}^\infty\frac{1}{2^n-1}\le\sum_{n=1}^\infty\frac{1}{2^{n-1}}=2,$$ which implies that $\displaystyle{\sum_{n=1}^\infty\frac{1}{2^n-1}}$ converges.
Let's use the integral test, which was what you thought was the way to go.
We consider $$I=\int_{1}^{\infty} \frac{1}{2^x -1 } \ dx.$$
Multiply top and bottom of integrand with $2^{-x}$ so that
$$I=\int_{1}^{\infty} \frac{2^{-x}}{1 -2^{-x} } \ dx.$$
Let $u=1-2^{-x}.$ Then $du = 2^{-x} \ln(2) \ dx.$
So $$I= \int_{\frac{1}{2}} ^{1} \frac{1}{\ln(2) u} \ du = \frac{\ln(1)-\ln \left(\frac{1}{2} \right)}{\ln(2)}= \frac{\ln(2)}{\ln(2)}=1.$$ So the integral converges, and so the sum in question converges too.
Compare $2^n-1$ to $n^\frac98$, which is smaller for all $n>1$. This means $\sum_0^\infty n^{-\frac98} > \sum_0^\infty \frac{1}{2^n-1}$ for all $n>1$. Since $\frac{1}{2^n-1}$ with $n=1$, and $\sum_0^\infty n^{-\frac98}$ is convergent (by the integral test), we know that since $\sum_0^\infty \frac{1}{2^n-1}$ is bigger for all $n>1$, it must be convergent too.
