Demonstrating a geometric series' convergence using partial sums

Question

I have this series $$\sum_{n=1}^{\infty}\left(\frac12\right)^{n-1}$$ and I need to demonstrate that it converges using partial sums.

Answer 1

Hint: The formula for the sum of a geometric series up to $n$ terms is $s_r= \frac{1-r^n}{1-r}$ where r is the common ratio. What is $\lim_{n\to\infty} s_{1/2}$?

Answer 2

Look at:

$$\sum_{n=0}^{N} r^n$$

Where $r \neq 1$. This is just a geometric sum and it can be evaluated Is there a name for the sum of increasing powers? to:

$$\frac{1-r^{N+1}}{1-r}$$

Use this to come up with a formula for:

$$\sum_{n=1}^{N} (\frac{1}{2})^{n-1}$$

Then look at what happens when $N \to \infty$

Answer 3

If you set $$ S_N = \sum_{n=0}^{N}\frac{1}{2^n} $$ you may easily check that $S_{N+1}=1+\frac{S_N}{2}$ and that $S_N+\frac{1}{2^N}=2$ (just think to the binary representation of $S_N$). Since the sequence $\{S_N\}_{N\geq 0}$ is increasing and bounded above, it is converging, and its limit has to fulfill $L=1+\frac{L}{2}$, from which $L=2$.

Answer 4

If $N \geq 1$, then $$ 1 - \bigg( \frac{1}{2} \bigg)^{N} = \bigg( 1 - \frac{1}{2} \bigg)\bigg[ 1 + \frac{1}{2} + \cdots + \bigg( \frac{1}{2} \bigg)^{N-1} \bigg]; $$ hence $$ \sum_{1}^{N}\bigg( \frac{1}{2} \bigg)^{n-1} = 2 - \bigg( \frac{1}{2} \bigg)^{N-1} \to 2 $$ as $N \to \infty$. You may try to prove that $2^{1-N} \to 0$ as $N \to \infty$ by an epsilon-argument.