Determine the sum of the given series:

1. Jul 22, 2014

ybhathena

1. The problem statement, all variables and given/known data

Sum starting from n=1 to infinity for the expression, (3/4^(n-2))

What the solutions manual has done is multiply the numerator and the denominator by 4.

12/(4^(n-1))

I don't know what they have done from here on:

12 / (1 - 1/4)

= 16

Can someone explain the solution? Thank you.

2. Jul 22, 2014

gopher_p

It's a geometric series.

3. Jul 22, 2014

ybhathena

I'm having trouble seeing how to use the fact that it is a geometric series in the problem. Could you explain further?

4. Jul 22, 2014

gopher_p

Do you have a formula for the sum of a geometric series?

5. Jul 22, 2014

ybhathena

I know for a geometric series the formula is Sn = (a(1-r^n)) / 1-r and for an infinite geometric series S = a/1-r

6. Jul 22, 2014

gopher_p

OK then ... does $\frac{12}{1-\frac{1}{4}}$ look like one of those forms?

7. Jul 22, 2014

ybhathena

Well the rate is 4, and the first value, a, is 1/4. Also 12 fits no where into any of these formulas so 12/(1-1/4) doesn't look like either of the two formulas for geometric series.

8. Jul 22, 2014

ybhathena

Oh wait they are treating the entire expression as a geometric series not just the denominator. Is that it? Give me a second to see then.

9. Jul 22, 2014

ybhathena

Ok I get it now I am finally getting 16 as an answer. My question is now why don't they divide 16 by 4 to get the original value to which the series converges to. (because in the beginning they multiplied the entire expression by 4?)

10. Jul 22, 2014

gopher_p

They multiplied both the numerator and denominator of the term inside the sum by 4, which is the same as multiplying the whole term inside the sum by 4/4=1. Multiplying by 1 doesn't change anything, so there is nothing to "undo". It goes $$\sum\limits_{n=1}^\infty\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty1\cdot\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty\frac{4}{4}\cdot\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty\frac{12}{4^{n-1}}=\sum\limits_{n=1}^\infty12\cdot\frac{1}{4^{n-1}}=\sum\limits_{n=1}^\infty12\cdot(\frac{1}{4})^{n-1}$$ with the formula for the geometric series applied to the RHS.
Everything's equal, so ...

11. Jul 22, 2014

ybhathena

Oh yes that is true the constant that is being multiplied is 1 not 4, so it won't change anything. Thank you for your help throughout this process; it is greatly appreciated!

12. Jul 22, 2014

HallsofIvy

Staff Emeritus
$$\sum_1^\infty \frac{3}{4^{n-2}}= \sum_1^{\infty} \frac{3}{16}\frac{1}{4^n}= \frac{3}{16}\sum_1^\infty \frac{1}{4^n}$$

BUT the formula for the sum of a geometric series: $$\sum_{n=0}^\infty a^n= \frac{1}{1- a}$$ starts with n= 0, not n= 1. There are two ways to handle that.
1) factor a "1/4" out of the series:
$$\frac{3}{64}\sum_{n=1}^\infty \frac{1}{4^{n-1}}= \frac{3}{64}\sum_{i= 0}^\infty \frac{1}{4^i}$$ where i= n-1.

2) Do the entire geometric series from 0 to infinity, then subtract off the n= 0 term:
$$\frac{3}{4}\sum_{n=0}^\infty \frac{1}{4^n}- \frac{3}{4}$$
since when n= 0, $$\frac{1}{4^n}= 1$$.

Last edited: Jul 22, 2014
13. Jul 22, 2014

gopher_p

Some texts give $\sum\limits_{n=1}^\infty ar^{n-1}=\frac{a}{1-r}$ as (one of) the formula(s) for the geometric series. Given the answer from the solutions manual, I'm guessing the topic creator is using one of those texts.