Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

While during physics I encountered a sum I couldn't evaluate: $$S= 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\cdots$$ Is there a particular formula for this sum and does it converges?

share|cite|improve this question

marked as duplicate by T. Bongers, sandwich, Hakim, RecklessReckoner, Jeel Shah Jul 5 '14 at 3:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Are you aware that an infinite series is a limit of its partial sums (the sum of the first $n$ terms, as $n \rightarrow \infty$)? If you sum the first 5 terms you get .6875, which might not seem so clear in terms of the trend, but the sum of the first 10 terms is .66601... and the sum of the first 20 terms is .66666603..., so the obvious conjecture to make is that $S = 2/3$. My point is not that this numerical experimentation is going to help you derive the exact answer, but it does make it quite clear what to guess the answer should be, and that's an important skill. – KCd Jul 4 '14 at 18:45
How did this appear in physics? If you don't mind me asking lol – Ethan Jul 4 '14 at 18:51

Hint: This is a geometric series with ratio $-\dfrac{1}{2}$.

Please see the wikipedia article on geometric series.

In general we have that $1+r+r^2+\ldots=\dfrac{1}{1-r}$ given $|r|<1$.

share|cite|improve this answer

We sum the geometric series:


share|cite|improve this answer
You've been busy here! ;-) – amWhy Jul 6 '14 at 12:05

Another easy way is to note that $2S=2-S$, whence $S=\frac{2}{3}$

share|cite|improve this answer

Using techniques that are frequently used by physicists: \begin{align*} 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots &= \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{4} - \frac{1}{8} \right) + \cdots \\ &= \frac{1}{2} + \frac{1}{8} + \cdots \\ &= \frac{1}{2} \left( 1 + \frac{1}{4} + \frac{1}{16} + \cdots \right) \\ &= \frac{1}{2} \frac{1}{1-\frac{1}{4}} \\ &= \frac{1}{2-\frac{1}{2}} \\ &= \frac{1}{3/2} \\ &= \frac{2}{3} \end{align*}

Note: There is no error in the parenthesization in the first line. We know the series converges (alternating series test), so every subsequence of the partial sums converges to the same thing, so this particular subsequence of the partial sums converges to the same thing.

share|cite|improve this answer

$$S=1 - \frac {1}{2} + \frac {1}{4} - \frac {1}{8} ...$$ Let this be (1)

Multilply both sides by $\frac {1}{2}$ we get

$$\frac{S}{2}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8} ...$$ and this be (2)

Now Add (1) and (2)

$$\frac{3}{2} \cdot S=1$$

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

share|cite|improve this answer

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