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 =)

  • 1
    $\begingroup$ are you sure that your term is correct? $\endgroup$ – Dr. Sonnhard Graubner May 25 '17 at 20:40
  • $\begingroup$ for your term i have got $$2^{-i-1} \left(\frac{1}{3} \left(1-(-2)^i\right)+2^{i-1}\right)$$ $\endgroup$ – Dr. Sonnhard Graubner May 25 '17 at 20:42
  • 1
    $\begingroup$ The right-hand term in the numerator is a geometric sum. Compute it and simplify. $\endgroup$ – user384138 May 25 '17 at 20:42
  • $\begingroup$ It looks like you mean $2^i$ instead of $2^{i-1}$ $\endgroup$ – B. Mehta May 25 '17 at 20:42
  • $\begingroup$ $$\sum_{j=0}^{i-1} (-1)^j2^j=-1+2-4+...=\dfrac{-1(1-(-2)^{i-1+1})}{1-(-2)}=\\\dfrac{(-2)^i-1}{3}$$ $\endgroup$ – Khosrotash May 25 '17 at 20:43

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: $$ \lim_{i\to \infty}a_{2i} = \frac{2}{3}, \qquad \lim_{i\to \infty}a_{2i-1} = \frac{1}{3}\cdot $$

  • $\begingroup$ Yes!! I'm working to prove that two sequences converge to $1/3$ and $2/3$ though I didn't know how. Your post is very helpful, thank you =) $\endgroup$ – T.S.Juve May 25 '17 at 21:03

$$\sum_{j=0}^{i-1} (-1)^j2^j=-1+2-4+...=\dfrac{-1(1-(-2)^{i-1+1})}{1-(-2)}=\\\dfrac{(-2)^i-1}{3}$$ $$\lim_{i\to \infty} \frac{2^{i-1} + \sum_{j=0}^{i-1} (-1)^j2^j}{2^{i+1}}=\\ \lim_{i\to \infty} \frac{2^{i-1} + \dfrac{(-2)^i-1}{3}}{2^{i+1}}$$can you go on ?

  • 1
    $\begingroup$ @OpenBall,it seems I confused these things,thank for correction $\endgroup$ – haqnatural May 25 '17 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.