How can I go about finding the limit of $$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{k = 1}^{\infty} \frac{1}{2^{k+1}}?$$ Could I use the absolute value theorem? I have a feeling it converges to $0$ but I am not sure.

  • 1
    Of the series or the sequence? – Sloan Jul 2 '15 at 17:26
  • sorry for the confusion. The third one and the limit of the series. – Lil Jul 2 '15 at 17:30
  • Divide every term by $2$ and see what happens. – Yves Daoust Jul 3 '15 at 12:28
  • The sequence converges to zero, but the sum of the terms converges to a positive value. Be sure you make a distinction between the values of the individual terms and the sum of the terms. – John Molokach Jul 3 '15 at 16:12

You know the limit of $1+\frac12 +\frac14+\frac18+\ldots$, don't you?

Your sequence is just like that, but without the $1$ and the $\frac12$.

Using the formula for the geometric series $$\sum\limits_{k=0}^\infty q^k=\frac{1}{1-q}$$ with $|q|<1$ we have: $$\sum\limits_{k=0}^\infty \frac{1}{2^{n+1}}=\sum\limits_{k=0}^\infty \left(\frac{1}{2}\right)^{n+1}=\sum\limits_{k=0}^\infty \frac{1}{2}\cdot \left(\frac{1}{2}\right)^n = \frac {1}{2}\cdot\sum\limits_{k=0}^\infty \left(\frac{1}{2}\right)^n = \frac {1}{2}\cdot \frac{1}{1-\frac{1}{2}}=\frac{1}{2}\cdot 2=1.$$

  • The problem was modified after you answered it. – N. F. Taussig Jul 3 '15 at 10:03
  • Well, if we start with $k=1$ we can easily obtain $$\sum\limits_{k=1}^\infty \frac {1}{2^{n+1}}=\sum\limits_{k=0}^\infty \frac{1}{2^{n+1}}-\frac{1}{2^{0+1}}=1-\frac{1}{2}=\frac {1}{2}.$$ – Hirshy Jul 3 '15 at 10:17

If your series begins at $\frac{1}{4}$, then it should be:

The sequence $a_n = \dfrac{1}{2^{n+2}}$ converges to $0$ as $n \to \infty$.

The series $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}=\frac{1}{4} \cdot \sum_{n=0}^{\infty} \frac{1}{2^{n}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$$ is the sum of a geometric sequence with common ratio $\frac{1}{2}$ and first term $\frac{1}{4}$. The series is convergent since the absolute value of the common ratio, $|r| = \frac{1}{2} < 1$. The sum evaluates to $$\frac{\frac{1}{4}}{1- \frac{1}{2}} = \frac{1}{2}$$ by using the general formula for the sum of a geometric series with common ratio $r$ and first term $a$, namely:$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$ as long as $|r| < 1$.

  • maybe I have the wrong formula.. my series is 1/4+1/8+1/16+1/32... – Lil Jul 2 '15 at 17:54
  • It begins at 1/4 not 1/2 as everyone is commenting. – Lil Jul 2 '15 at 17:54
  • 1
    Then your series is $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}$$ since the sum of a geometric series starts from $n=0$ by convention. :-) – Zain Patel Jul 2 '15 at 17:57
  • I don't understand why it would be 1/2^n+2 because when n=1 it would become 1/2^3 which is 1/8 but my first term is 1/4...? – Lil Jul 2 '15 at 17:58
  • doesn't the sum of a geometric series begin at 1? or no? – Lil Jul 2 '15 at 17:58

Use $$\sum_{n=0}^\infty a r^n=\frac{a}{1-r},\quad |r|<1.$$ Take $r=\frac12$ and $a=\frac14$ since the first term of your series is $\frac14$ and the common ratio of your Geometric progression is $\frac12.$

  • The problem was modified after you answered it. – N. F. Taussig Jul 3 '15 at 10:03

I will add an indefinite version of the sum. In difference calculus we have that the function $2^n$ is the analogous of $e^x$ on ordinary differential calculus.

We have that $\sum 2^n \delta n=2^n$ (you can see it by yourself if you do the difference and see that it doesnt change, i.e. $\Delta 2^n=2^{n+1}-2^n=2^n$).

In your sum, doesnt taking limits, we have that

$$\sum 2^{-n-1}\delta n=-2^{-n}$$

If we take limits then

$$\sum_{n=1}^{\infty} 2^{-n-1}=-2^{-n}\bigg\lvert_{1}^{\infty}=-2^{-\infty}+2^{-1}=\frac{1}{2}$$

Take care with notation because

$$\sum_{k\ge 0}2^k\ne\sum_{k\ge 0}2^n$$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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