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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 =)

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

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

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

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