# Geometric Series Question

1. Apr 14, 2013

### brojas7

$\sum^{\infty}_{k=4}$ $\frac{1}{5^{k}}$

ar^n=a/(1-r)

Here is what I did:

a=5
r=1/625

(1/5)/(1-1/625)=(1/5)/(624/625)=625/3120=125/624

Where am I going wrong?

2. Apr 14, 2013

### Dick

You've got a and r wrong. a is the first term in the series and r is the common ratio. Try again.

3. Apr 14, 2013

### Fredrik

Staff Emeritus
Can you rewrite the sum in the form
$$\frac{1}{5^4}\sum_{i=0}^\infty(\text{something})_i?$$ This isn't absolutely necessary, but if you do, you can use the version of the formula with a=1.

4. Apr 14, 2013

### brojas7

i meant a to be 1/5...
the first term was 1/625

5. Apr 14, 2013

### brojas7

If I do it that way,r would equal 1/5 right?

$$\frac{1}{5^4}\sum_{i=0}^\infty(\text{(1/5)^n})?$$

6. Apr 14, 2013

### Fredrik

Staff Emeritus
Yes, assuming that you meant what I hope you meant when you wrote ar^n=a/(1-r), which doesn't really make sense.

7. Apr 14, 2013

### brojas7

It worked, thank you so much

8. Apr 14, 2013

### SithsNGiggles

The sum of an infinite geometric series of the form $\displaystyle\sum_{n=0}^\infty ar^n=\dfrac{a}{1-r}$, for $|r|<1$.

Your given series starts off with $n=4$. Think you can rewrite the series to make it start with $n=0$?

9. Apr 14, 2013

### brojas7

I didn't know in order for the formula to work I had to make n=0. Thanks