# How can I compute $\sum_{n=0}^{\infty} 0.6^n$? [duplicate]

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.

## marked as duplicate by user147263, Surb, Rolf Hoyer, Aaron Maroja, kennytmApr 22 '15 at 16:39

• The general formula for computation of a geometric series with first term $a$ and common ratio $r$, when there are $n$ terms in the series is,$$S=a\cdot\frac{r^n-1}{r-1}$$ – Prasun Biswas Apr 22 '15 at 10:47
• When the sum has $n\to\infty$, you can take the limit of the sum as $n\to\infty$. Note that the infinite sum diverges if $|r|\geq 1$ – Prasun Biswas Apr 22 '15 at 10:48
• or Infinite Geometric Series Formula Derivation if you prefer – user147263 Apr 22 '15 at 11:52

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.

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:

$$\sum_{k=0}^{\infty}0.6^{k}=\frac{1}{1-0.6}=\frac{5}{2}$$

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 = $\frac{1}{1-0,6} = 2,5$.

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 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

$S_n(r - 1) = a(r^{n+1} - r^0)$

$S_n = \frac{a(r^{n+1} - 1)}{(r - 1)}$