Sum of a Geometric series Approximately 2? or exactly 2? (1+1/2+1/4+...) My question is not only asking specifically about the answer, but how to get there. I've always had the notion that this infinite series when summed up would be very close to 2, but not exactly 2. I'm not sure if this is simple to answer or hard. I'm curious about the properties of infinity, but may not have the knowledge to delve into it. I've studied up to Calculus 2 in College if that helps.
 A: Here's a way to think of it without limits.
$S \;\,= 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\ldots$
$2S = 2+1+\frac{1}{2}+\frac{1}{4}\ldots$
Subtract and all of the terms cancel except for the first one in $2S$ because there are an infinite amount of terms.
$2S-S=2$
$S=2$
The exact value of the infinite sum is 2.
A: The value of an infinite series is defined as a limit.   It either is a value, or does not exist. 
In this case (presuming you meant this series), the sequence converges so the series exists and its value is exactly $2$.
$$\begin{align} \sum_{k=0}^{\infty} \frac 1 {2^k} ~=~ & \lim_{n\to\infty} ~\sum_{k=0}^n \frac 1 {2^k} & = 1+\tfrac 12 +\tfrac 14 +\cdots \\[2ex] ~=~& \lim_{n\to\infty} \dfrac{2-2^{-n}}{2-1} \\[2ex]~=~& 2\end{align}$$
It is not the case that "... but the series never gets to $2$" or anything like that.    The infinite series is not "going" anywhere.   It is a limit of a sequence of finite series, and that limit is a constant: $2$.
A: First off, you seem to be writing the harmonic series,
$$\sum_{k=1}^{\infty}\frac1k$$
which diverges.
The sum of a series is defined as the limit of the sequence of partial sums, so the geometric series would be
$$\lim_{N\rightarrow\infty}\sum_{k=0}^Nr^k$$
The partial sum are usually derived by defining
$$S_N=\sum_{k=0}^Nr^k$$
So that $$\begin{align}(r-1)S_N&=\sum_{k=0}^Nr^{k+1}-\sum_{k=0}^Nr^k=\sum_{k=1}^{N+1}r^k-\sum_{k=0}^Nr^k\\
&=\sum_{k=1}^Nr^k+r^{N+1}-\sum_{k=1}^Nr^k-1=r^{N+1}-1\end{align}$$
So
$$S_N=\sum_{k=0}^Nr^k=\frac{r^{N+1}-1}{r-1}$$
Now, if $|r|<1$ this limit
$$\lim_{N\rightarrow\infty}\sum_{k=0}^Nr^k=\lim_{N\rightarrow\infty}\frac{r^{N+1}-1}{r-1}=\frac1{1-r}$$
Exists, so that is the value of the geometric series exactly. If $|r|\ge1$, of course the limit does not exist so the geometric series doesn't have a real value.
A: Assume that the sum in question is actually the geometric
$$1+\frac12+\frac14+\frac18+\dots=\sum_{i=0}^{\infty}2^{-i}$$
If we compute a partial sum at
$$S(n)=\sum_{i =0}^{n}2^{-i}$$
we find that we can evaluate $2-S(n)=2^{-n}$.  So each time we increase $n$ by one, we close the distance from the amount we have summed up to the limit amount of $2$ by cutting that distance in half.  Taking the limit as $n\to\infty$, we see that $2-S(n)=2^{-n}\to 0$, and therefore $S(n)\to 2$.  The sum only reaches $2$ in the limit; at every partial sum along the way, there is still a non-zero distance between the sum and $2$.
