What is the limit to infinity of this function with summation? I need someone with knowledge of limits to infinity and limits of summations to please work out the following:
$$ \lim_{i\to \infty} \frac{2^{i} + \sum_{j=0}^{i-1} (-1)^j2^j}{2^{i+1}} $$
For context, I want to determine if the following sequence approaches $\frac{2}{3}$:
$$ \frac{1}{2}, \frac{3}{4}, \frac{5}{8}, \frac{11}{16}, \frac{21}{32}, \frac{43}{64}, \frac{85}{128}, \frac{171}{256}, \frac{341}{512}, . . . $$
Thank you =)
 A: Your sequence does not converge. Firstly $\sum_{j=0}^{i-1} (-1)^j 2^j$ is sum of geometric series with quotient -2 and is equal to $\frac{1}{3} (1-(-2)^i)$. 
We have
$$ \lim_{i\to \infty} \frac{2^{i} + \sum_{j=0}^{i-1} (-1)^j2^j}{2^{i+1}} = 
\lim_{i\to \infty} \frac{2^{i}}{2^{i+1}} + \lim_{i\to \infty} \frac{ 1-(-2)^{i}}{3\times 2^{i+1}} = \frac{1}{2} + \lim_{i\to \infty} \frac{ 1}{3\times 2^{i+1}}+ \lim_{i\to \infty} \frac{ -(-2)^{i}}{3\times 2^{i+1}} = \frac{1}{2} -\frac{1}{6} \lim_{i\to \infty} (-1)^i$$
Since the last does not converge, the limit does not converge. 
Actually the sequence with $i-$th term $a_i = \frac{1}{2} -\frac{1}{6}(-1)^i$ has two subsequences, that converge to $1/3$ and $2/3$, respectively:
$$
\lim_{i\to \infty}a_{2i} = \frac{2}{3}, \qquad \lim_{i\to \infty}a_{2i-1} = \frac{1}{3}\cdot
$$
A: $$\sum_{j=0}^{i-1} (-1)^j2^j=-1+2-4+...=\dfrac{-1(1-(-2)^{i-1+1})}{1-(-2)}=\\\dfrac{(-2)^i-1}{3}$$
$$\lim_{i\to \infty} \frac{2^{i-1} + \sum_{j=0}^{i-1} (-1)^j2^j}{2^{i+1}}=\\
\lim_{i\to \infty} \frac{2^{i-1} + \dfrac{(-2)^i-1}{3}}{2^{i+1}}$$can you go on  ?
