What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$

Question

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

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

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

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

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!!

Answer 1

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

Answer 2

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.

Answer 3

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...

Answer 4

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.

Answer 5

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.