Determine the sum of the given series:

[ybhathena]

Homework Statement

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.

 

[gopher_p]

It's a geometric series.

 

[ybhathena]

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

 

[gopher_p]

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

 

[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

 

[gopher_p]

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

 

[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.

 

[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.

 

[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?)

 

[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 ...

 

[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!

 

[HallsofIvy]

\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.

 

[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.