limit when $\lim \limits _{n \to \infty} \sum _{i=0} ^{n} \frac{1}{2^i}$

Question

limit when $$\lim \limits _{n \to \infty} \sum _{i=0} ^{n} \frac{1}{2^i}$$ I intuitively know that the limit is 2 and that a series is convergent, but I do not know how to arrive at the result

Answer 1

Let $$S_n=\sum_{i=0}^n\frac{1}{2^i}$$

then

$$S_{n+1}=1 +\frac 12 S_n$$

The limit $L$ should satisfy

$$L=1+\frac L2$$

Answer 2

If you want to do it without invoking the magic words "geometric series", note that if $S = \sum_{i=0}^n \frac{1}{2^i}$ then $$\frac{1}{2}S = \sum_{i=1}^{n+1} \frac{1}{2^i} = \underbrace{\sum_{i=0}^n \frac{1}{2^i}}_{S} + \frac{1}{2^{n+1}} - \frac{1}{2^0}$$ which you can rearrange.

Answer 3

Use the formula for the sum of a geometric series:

$$\sum_{i=0}^n a_0\cdot r^i=\frac{a_0(1-r^{n+1})}{1-r}$$

Here, $a_0=1$ and $r=\frac 1 2$, so we have:

$$\sum_{i=0}^n \frac{1}{2^i}=\frac{1\cdot \left(1-\left(\frac{1}{2}\right)^{n+1}\right)}{1-\frac{1}{2}}=2\left(1-\frac{1}{2^{n+1}}\right)$$

Now, we need to find:

$$\lim_{n \to \infty} 2\left(1-\frac{1}{2^{n+1}}\right)$$

Since $\frac{1}{2^{n+1}} \to 0$ as $n \to \infty$, we get:

$$\lim_{n \to \infty} 2\left(1-\frac{1}{2^{n+1}}\right)=2(1-0)=2$$