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The answer I got by hand is not the same to the one I found using a spreadsheet.

$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$


$\displaystyle \frac{1}{4}S = \hspace{8.5pt} \frac{1}{4} + \frac{3}{16} + \frac{7}{64} + \frac{15}{256} + \ldots$

$\displaystyle \frac{3}{4}S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \qquad \leftarrow S- \frac{1}{4}S$

For the Infinite Sum on the RHS $\displaystyle \left(S = \frac{a}{1-r}\right)$:

$\displaystyle a = 1$

$\displaystyle r = \frac{1}{2}$

Then

$\displaystyle \frac{3}{4}S = \frac{1}{1-\frac{1}{2}}$

$\displaystyle \frac{3}{4}S = 2$

$\displaystyle S = \frac{8}{3}$

Using Excel the answer is $\displaystyle \frac{5}{3}$, but I don't know where is the issue.

Thanks!!

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    $\begingroup$ try to learn some python, is better than excel! $\endgroup$
    – L F
    May 13, 2015 at 21:01
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    $\begingroup$ Why is $\frac34 S $ is what you wrote up there,? $\endgroup$
    – Atvin
    May 13, 2015 at 21:01
  • $\begingroup$ @Atvin I subtracted the series $\frac{1}{4}S$ from $S$. Thus $S - \frac{1}{4}S = \frac{3}{4}S$. $\endgroup$
    – Joqsan
    May 13, 2015 at 21:03
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    $\begingroup$ It is highly likely that you forgot the first term $1$ in Excel. $\endgroup$
    – user65203
    May 13, 2015 at 21:11
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    $\begingroup$ @RobArthan Take it easy. I just did the thing up to 50 terms (in Excel is a good approximation to infinity). As Yves Daoust says, I forgot to add the first term: $1$. $\endgroup$
    – Joqsan
    May 13, 2015 at 21:22

5 Answers 5

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Hint: Notice that $$ S = \sum_{i=0}^\infty \frac{2^{i+1}-1}{2^{2i}} $$ and distribute.

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Your answer seems to be correct. We see that Excel's answer is wrong by adding just the first 2 terms in the series, $1, \frac 34$ and we see that $1.75>1.66..$, so we can see that Excel's answer is wrong.

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We can pattern match this to a difference of geometric series. The sum in question is $$\begin{align} 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} = 1+\sum_{k=1}^\infty \frac{2^{k+1}-1}{4^{k}} \\ = 1+\sum_{k=1}^\infty \frac{2^{k+1}}{4^{k}}-\sum_{k=1}^\infty \frac{1}{4^{k}}\end{align}$$ Now use geometric series...

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  • $\begingroup$ Better to write answer without ambiguity...+1 $\endgroup$
    – k1.M
    May 13, 2015 at 21:10
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At first, note that $S=\dfrac{2^1-1}{4^{1-1}} + \dfrac{2^2-1}{4^{2-1}}+ \dfrac{2^3-1}{4^{3-1}}+\dfrac{2^4-1}{4^{4-1}} + \dfrac{2^5-1}{4^{5-1}}+\ldots $

So: $$S=\displaystyle\sum_{k=1}^{\infty}\dfrac{2^k-1}{4^{k-1}}?$$

$$S=\displaystyle\sum_{k=1}^{\infty}\dfrac{2^k}{4^{k-1}}-\dfrac{1}{4^{k-1}}$$ $$S=\displaystyle\sum_{k=1}^{\infty}\dfrac{4}{2^k}-\dfrac{1}{4^{k-1}}$$ And you can go on since here.

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    $\begingroup$ That sum is not correct $\endgroup$ May 13, 2015 at 21:01
  • $\begingroup$ $\sum_{k=1}^{\infty}\dfrac{2^k-1}{2^{k+1}}=1/4+3/8+\dots$ $\endgroup$ May 13, 2015 at 21:02
  • $\begingroup$ Hey downvoter man, I improve my answer 2 hours ago, you are so fast for somethings but slow for others. $\endgroup$
    – L F
    May 13, 2015 at 23:48
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When you write "Using Excel the answer is $\frac{5}{3}$", I think what you mean is that you have constructed a model of the problem in Excel that suggests the answer is $\frac{5}{3}$. Your Excel model of the problem is wrong (as calculation by hand of the first few terms of of the sum shows). So the issue is in your Excel model. Debugging your Excel model is off topic for MSE.

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