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.
 A: Use comparison:
$$\sum_{k=0}^{\infty}{\frac{k}{4^{k-1}}}<\sum_{k=0}^{\infty}\frac{2^k}{4^{k-1}}=4\sum_{k=0}^{\infty}\bigg(\frac{2}{4}\bigg)^k$$
A: With a few tools that the OP probably doesn't have yet, we can also find the actual limit:
It is well known that $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n \text{ when } |x|<1$$
Differentiating this, we get
$$\frac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1}, $$
which is just your series if we set $x=1/4$.
Since $|1/4|<1$ and differentiation doesn't change the radius of convergence, your series converges, and the limit is $\frac{1}{(1-1/4)^2} = \frac{16}{9}$.
A: the series $\sum (x_n-x_{n-1})$ converges, because
$$x_n-x_{n-1}=\frac n{4^{n-1}}\leq\frac1{2^{n-1}}$$
A: $\int_{1}^{\infty}\frac{n}{4^{n-1}}dn=\frac{2ln(2)+1}{4ln(2)^2}$
and
$\lim_{t \to \infty}\int_{1}^t\frac{n}{4^{n-1}}dn=\frac{2ln(2)+1}{4ln(2)^2}$
(Integral test)
A: By induction the partial sums are $\frac{16*4^{n-1} - 3n -4}{9*4^{n-1}}$, which clearly approaches $\frac{16}{9}$ as $n$ goes to infinity.
