Help finding the limit of this series $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots$

Question

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.

Answer 1

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 $\frac12.$

Answer 2

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

Answer 3

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 $|r| < 1$.

Answer 4

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 $\frac12$.

Answer 5

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

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