estimation of a series $3^n/( 4^n -1 )$

Question

I am trying to show that the series $$ \sum_{i=0}^\infty \frac{3^i}{4^i-1}$$ is convergent, but do not see how to get rid of the one in order to get a bigger series. Thanks for helping.

Answer 1

You can use the note in the comments that $4^i-1>4^{i-1}$, and thus $$\frac{3^i}{4^{i}-1}<\frac{3^i}{4^{i-1}}$$ so $$\sum \frac{3^i}{4^{i}-1}<\sum \frac{3^i}{4^{i-1}}<\infty$$

In general, if one series looks like another you can apply the limit comparison test. This is often much easier when finding an explicit bound is tricky.

$$\lim_{i\to\infty} \frac{3^i}{4^{i}-1} \frac{4^i}{3^{i}}=\lim_{i\to\infty} \frac{4^i}{4^{i}-1} =1$$