How can I compute $\sum_{n=0}^{\infty} 0.6^n$? I am a computing teacher and just helping out some students with a math question. They have been asked to calculate the following:
$$\sum_{n=0}^{\infty} 0.6^n$$
I am intrigued as to how one gets to the answer - i.e. the logic and process behind it if possible. 
 A: Note that
$$
(x^{n-1}+...+x+1)(x-1)=x^n-1
$$
so
$$
\sum_{i=0}^{n-1} x^i=\frac{x^n-1}{x-1}
$$

You have $x=0.6$, so
$$
\sum_{i=0}^{n-1} 0.6^i=\frac{0.6^n-1}{0.6-1}\longrightarrow\frac{0-1}{0.6-1}=2.5\quad\text{for }n\to\infty
$$
As others have mentioned, this is known as a geometric series and is convergent for $|x|<1$, which should be evident from the expression $S_{n-1}=\frac{x^n-1}{x-1}$ for the partial sums.
A: This type of summation is known as a geometric series, and so we have the following identity:
$$\sum_{k=0}^{\infty}\alpha^{k} = \frac{1}{1-\alpha} \quad \iff \quad |\alpha|<1$$
So in this case, $\alpha = 0.6 < 1$ and so:
$$\sum_{k=0}^{\infty}0.6^{k}=\frac{1}{1-0.6}=\frac{5}{2}$$
A: This is a geometric series in the following form: $\rightarrow \frac{6}{10} + \frac{36}{100}+...$. For arbitrary $n$, the $a_n = \frac{6^n}{10^n}$.
You can see that the quotient is $\frac{6}{10}$, which is smaller than $1$(in that case, the sum is finite), so we can use the formula to calculate the sum:
Sum = $\frac{1}{1-0,6} = 2,5$.
A: This is a geometric series with parameter $0.6$. For any $0<r<1$, you can get a formula for the partial sums
$$
S_n=\sum_{k=0}^n r^k.
$$
Multiplying both sides by $r$, you get
$$
rS_n=\sum_{k=0}^nr^{k+1}
$$
The difference of those expressions is then
$$S_n-rS_n=r^0-r^{n+1}=1-r^{n+1}.$$
Then you can conclude that 
$$
S_n(1-r)=1-r^{n+1} \Leftrightarrow S_n=\frac{1-r^{n+1}}{1-r}.
$$
Letting $n\to \infty$, as $0<r<1$, you get
$$
\sum_{k=0}^{\infty}r^k=\lim_{n\to \infty} S_n=\lim_{n\to \infty}\frac{1-r^{n+1}}{1-r}= \frac{1}{1-r}.
$$
In this particular case the answer would be $1/0.4=2.5$.
A: A geometric progression is the sum of s series with elements of the form
$T_n = a r^n$ 
for each term in the series.
The sum of the series is
$S_n = \sum_0^\infty{T_n} = ar^0 + ar^1 + ar^2 + ... + ar^n$
Consider multiplying by $r$
$S_n r = ar^1+ar^2 + ... + ar^{n+1}$
$S_nr - S_n = ar^{n+1} - ar^0 $ 
because all the other terms cancel out
$S_n(r - 1) = a(r^{n+1} - r^0) $ 
$S_n = \frac{a(r^{n+1} - 1)}{(r - 1)} $ 
