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