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

As a follow-up to a previous question, I'd like to know if

$$\sum_{n=1}^\infty\frac{1}{2 ^ n}$$

is convergent? If yes, then what does it converge to?

Just as a sidenote - the whole issue started with reading about Zeno's Turtle Paradox which made me curious about the underlying mathematics.

share|cite|improve this question
Sure, the terms go to $0$ real fast. – André Nicolas Jan 23 '12 at 19:46
up vote 10 down vote accepted

This series is a geometric series and converges, see the link in the comments.

But, I'll offer the standard proof that this particular series converges here:

A series is defined to converge to $L$ if and only if the sequence of its partial sums converges to $L$:

The infinite series $ \sum\limits_{n=1}^\infty a_n$ converges to $L$ if and only if the sequence of partial sums $\{S_n\}$, defined by $$S_n=\sum\limits_{m=1}^n a_m =a_1+a_2+a_3+\cdots+a_n,$$ converges to $L$.

If you consider the sum of the first $n$ terms, $S_n$, for $\sum\limits_{n=1}^\infty {1\over 2^n}$, then $$S_n=\sum\limits_{m=1}^n{1\over2^m}.$$ Writing out this sum explicitly gives $$ S_n={1\over 2}+{1\over 2^2}+\cdots+{1\over 2^n}. $$ Multiply both sides of the above by $1\over2$: $$ {1\over 2} S_n = {1\over 2^2}+{1\over 2^3}+\cdots+{1\over 2^{n+1}} $$ Now add $1\over2$ to both sides: $$ {1\over 2} S_n+{1\over2} =\underbrace{{1\over 2}+ {1\over 2^2}+{1\over 2^3}+\cdots+ +{1\over 2^n}}_{S_n}+{1\over 2^{n+1}} $$ Then: $$ {1\over 2} S_n+{1\over2} =S_n+{1\over 2^{n+1}}. $$ Solving the above for $S_n$ gives: $$ {1\over2}-{1\over 2^{n+1}}={1\over2}S_n $$ $$ S_n=1-{1\over 2^n}. $$ From the above, we can see that as $n\rightarrow\infty$, $S_n\rightarrow1$. So $\sum\limits_{n=1}^\infty {1\over 2^n}=1$.

Or, consider the partition of the unit square:

enter image description here

share|cite|improve this answer
+1, very nice wordless proof at the bottom! More nice proofs at – lhf Jan 23 '12 at 13:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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