Find lim: $x_n=\dfrac{1+\frac12+...+\frac1{2^n}}{1+\frac14+...+\frac1{4^n}}$ Find limit: $x_n=\dfrac{1+\frac12+...+\frac1{2^n}}{1+\frac14+...+\frac1{4^n}}$ as $n \rightarrow \infty$ My "intuition" says that it should be $\frac34$ but I don't know how to proove it with rigour.

By induction you can prove the following formula for the sum of terms of a geometric progression: $$a + ar + ar^2 + ... + ar^n = {a - ar^{n+1} \over 1 - r}$$ So you can apply this in the numerator with $a = 1, r = {1 \over 2}$, and in the denominator with $a = 1, r = {1 \over 4}$. You obtain $$x_n = {1 - {1 \over 2^{n+1}} \over 1 - {1 \over 2}} \times {1 - {1 \over 4} \over 1 - {1 \over 4^{n+1}}}$$ $$= {3 \over 2} {1 - {1 \over 2^{n+1}} \over 1 - {1 \over 4^{n+1}}}$$ Now take limits as $n \rightarrow \infty$.