# Thread: Finding the value of infinite sums

1. ## Finding the value of infinite sums

Okay, I've got this example in my notes and I don't understand it. My notes say...

Find $\displaystyle \sum_{k=0}^{\infty}3^{-k}$.
Step 1 Find $\displaystyle S_{1}$, $\displaystyle S_{2}$...
Step 2 Find the limit.

$\displaystyle S_{1}=\frac{1}{1\times2}=1-\frac{1}{2}$

But I don't understand that last bit. I know it's true (obviously) but why write it like that?

2. It's a simple geometric series. Other than that, I would not know why you wrote what you did.

$\displaystyle \sum_{k=0}^{\infty}\frac{1}{3^{k}}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$

3. $\displaystyle S_n=\frac{1-3^{-(n+1)}}{1-3^{-1}}=\frac{3}{2}(1-\frac{1}{3^{n+1}})$.

4. Originally Posted by chella182
Okay, I've got this example in my notes and I don't understand it. My notes say...

Find $\displaystyle \sum_{k=0}^{\infty}3^{-k}$.
Step 1 Find $\displaystyle S_{1}$, $\displaystyle S_{2}$...
Step 2 Find the limit.

$\displaystyle S_{1}=\frac{1}{1\times2}=1-\frac{1}{2}$

But I don't understand that last bit. I know it's true (obviously) but why write it like that?

I would do ths problem by looking at the first 3 terms in the series

From

$\displaystyle \sum_{k=0}^{\infty}3^{-k}$

term 1 $\displaystyle t_1$ for n=0 is $\displaystyle 3^{-0} = 1$

term 2 $\displaystyle t_2$ for n=1 is $\displaystyle 3^{-1} = \frac{1}{3}$

term 3 $\displaystyle t_3$ for n=2 is $\displaystyle 3^{-2} = \frac{1}{9}$

giving the series $\displaystyle {1 ,\frac{1}{3},\frac{1}{9}}\cdots$

From here I would find r = $\displaystyle \frac{t_3}{t_2} = \frac{t_2}{t_1} = \frac{1}{3}$

and 'a' the first term = 1

From here

$\displaystyle \lim_{n\to\infty} \frac{1-r^{n+1}}{1-r}$

and for the sum of infinate series

$\displaystyle S_{\infty} \frac{a}{1-r}$