calculus - Prove that $x_n=1+frac{2}{4}+frac{3}{16}+...+frac{n}{4^{n-1}}$ converges

so i have got a sequence $$x_n=1+\frac{2}{4}+\frac{3}{16}+\frac{4}{64}+...+\frac{n}{4^{n-1}}$$
and i have to prove that it actually converges to some point, just by looking at it, it is clear to me that it does converge, if i would take its limit $$\lim_{n \to \infty}\sum_{k=0}^{n}{\frac{n}{4^{n-1}}}$$

as n increases the numerator becomes actually less than the
denominator, from that point it would converge, but how would i prove it.

Answer

Use comparison:

$$\sum_{k=0}^{\infty}{\frac{k}{4^{k-1}}}<\sum_{k=0}^{\infty}\frac{2^k}{4^{k-1}}=4\sum_{k=0}^{\infty}\bigg(\frac{2}{4}\bigg)^k$$