# Geometric Series Question

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

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?

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

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

Fredrik
Staff Emeritus
Gold Member
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.

You've got a and r wrong. a is the first term in the series and r is the common ratio. Try again.
i meant a to be 1/5...
the first term was 1/625

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

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

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

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##?

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