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I noticed that $$\large{2x-1 = \frac{x}{2} + \frac{\frac{x}{2}}{2} + \frac{\frac{\frac x2}2}2} + \cdots$$ until the output of one of the steps in the pattern equals $1$. Or in other words $2x-1$ is equal to the sum of $\large{\frac{x}{2}}$ plus that result divided by two plus that result divided by two and so forth.

How do I notate this with a Sigma? Or would I not use Sigma and if so what is the best way to notate this? I cannot figure it out; any help will be appreciated!

Alice Ryhl
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Nick Suwyn
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  • Maybe you do not know that if $|r|\lt 1$ then the sum of the infinite geometric series $a+ar+ar^2+\cdots$ is $\frac{a}{1-r}$. That gives sum $x$. More simply, your sum is $x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)$, and the inner sum is $1$. – André Nicolas May 15 '15 at 21:17
  • Im not sure what sum you mean, but $\sum_{k=1}^{\infty} \frac{x}{2^k}=x$-not $2x-1$ – callculus42 May 15 '15 at 21:27
  • Thank you, both of you; you definitely answered my question and cleared up my mistake! – Nick Suwyn May 15 '15 at 21:45

2 Answers2

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Each step of your formula adds another division by two, this is the same as multiplying with $\frac12$. If you want to multiply a number with something a set number of times, you use exponents. This gives a formula for each component in the sum. $$ x\cdot\left(\frac12\right)^n $$ This means that the sigma notation of your sum would be: $$ \sum_{n=1}^\infty x\cdot\left(\frac12\right)^n=x\sum_{n=1}^\infty\left(\frac12\right)^n $$ The value of that infinite sum is $1$, so it turns out that the above expression doesn't evaluate to $2x-1$, instead: $$ \sum_{n=1}^\infty x\cdot\left(\frac12\right)^n=x $$

To evaulate the sum, we define $S_N$ to be $$ S_N=\sum_{n=1}^N\left(\frac12\right)^n=\frac12+\frac14+\cdots+\frac1{2^N} $$ Notice that the value of $\frac12S_N$ is given as below: $$ \frac12S_N=\sum_{n=2}^{N+1}\left(\frac12\right)^n=\frac14+\frac18+\cdots+\frac1{2^{N+1}} $$ Subtracting these from each other, and arranging the sum the following way gives: $$ S_N-\frac12S_N=\frac12+\frac14-\frac14+\cdots+\frac1{2^N}-\frac1{2^N}-\frac1{2^{N+1}} $$ Canceling terms on the right side and simplying the left side gives $$ \frac12S_N=\frac12-\frac1{2^{N+1}} $$ Multiply by 2 and join the fractions $$ S_N=\frac{2^N-1}{2^N}=1-2^{-N} $$ To evaluate the infinte sum, simply take the limit as $N\to\infty$, since $2^{-N}\to0$ as $N\to\infty$ we have the limit as $$ \sum_{n=1}^\infty\left(\frac12\right)^n=\lim_{N\to\infty}S_N=1 $$

Alice Ryhl
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The partial sum of this geometric series is $\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}$, if you start at k=0. Thus in your case it is $\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}-1$, because $2^0=1$.

$=\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}-\frac{1-\frac{1}{2}}{1-\frac{1}{2}}$=$\frac{\frac{1}{2}-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}$

If $n \to \infty$ then $\lim_{n \to \infty} \frac{\frac{1}{2}-\left( \frac{1}{2} \right) ^n}{\frac{1}{2}}=\frac{\frac{1}{2}-0}{\frac{1}{2}}=1$.

Consider that $\lim_{n \to \infty} a^n=0$, if $|a|<1$.

callculus42
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