Question

evaluate the infinite series

Answer 1

\[\sum_{n=1}^{\infty}3(\frac{ 1 }{ 4 })^{n-1}\]

Answer 2

so you need the limiting sum...

substitute n = 1 to find the 1st term...a

the common ratio is 1/4

then the limiting sum is \[S_{\infty} = \frac{a}{1 - r}\]

hope it helps

Answer 3

i still have no clue how to do it

Answer 4

This is the coolest of all series : The Geometric Series

Answer 5

Expand it and look at the terms may be :

3 + 3/4 + 3/4^2 + 3/4^3 + ...

Answer 6

See any pattern ?

Answer 7

It would be easy if you saw the pattern on your own
give it a try, Im sure you'll get it

Answer 8

yeah but this has a infinity sign

Answer 9

yeah its weird that adding infinity terms gives you something finite, but do you see any pattern in the terms ?

3 + 3/4 + 3/4^2 + 3/4^3 + ...

Answer 10

yes i see it

Answer 11

What is it, can you explain whatever the pattern that you see ?

Answer 12

@shaeelynn ....u get it ?

Answer 13

no

Answer 14

If you write out the series you will see that it takes the form of an infinite geometric series
You should be able to go from the sigma notation to this (If you can't, remember on the sigma it has two numbers above and below. The lower number n = ? tells you the first integer to substitute into the formula on the right of the sigma. The upper number tells you the last number to substitute. As this is an infinite series, we just write the first few numbers and then leave a gap, and write the last number (for infinity, just put n)

\[3 + \frac{ 3 }{ 4 } +\frac{ 3 }{ 4^2 } + \frac{ 3 }{ 4^3 } + ... +\frac{ 3 }{ 4^n } (n \rightarrow \infty) = 3 + \frac{ 3 }{ 4 } +\frac{ 3 }{ 16 } + \frac{ 3 }{ 64 } + ... +\frac{ 3 }{ 4^n }\]

If you look closely, each new term is generated by multiplying its previous term by 1/4
(or dividing each previous term by 4)

Now if the sum of the infinite geometric series CONVERGES, this means we can calculate the sum to be a finite number, also known as a limiting sum.

This limiting sum, S, can be found using this formula
\[S_\infty = \frac{ a }{ 1-r }\]
where a is the first term in the series and r is the common ratio.

I'll let you find out what a and r are from the above series, just plug that in.

Answer 15

as another hint you can try adding say 5-10 terms from the series on your calculator and see if the result starts to plateau towards a particular number. Of course it's imperative you know the formula for limiting sum, which saves you a lot of time.

Answer 16

i don't understand this i need step by step directions on how to do it

Answer 17

dont worry about the infinity sign. think of it like its telling u that it is quite a long series that goes on and on. its not a number and u dont have to use it in ur solution

Answer 18

what ur solution requires is that u tell what happens if u keep on adding those terms

Answer 19

if u observe carefully u will notice that each progressive term is getting smaller

Answer 20

if u keep adding these terms first it will affect ur sum significantly

Answer 21

but eventually terms will get so small that the sum will lean towards a single number

Answer 22

and that number is given by \[S \infty = \frac{ a }{ 1 -r }\]

Answer 23

S∞ is the number u require. 'a' is the first term of series, while 'r' is the common ratio.

Answer 24

okay so is 4 the common ratio?

Answer 25

The common ratio is what you multiply each term by.

Answer 26

u find the common ratio by: \[\frac{ U _{n+1} }{U _{n}}\]

Answer 27

i used U to refer to any term of the series

Answer 28

so common ratio can by found by dividing any term by its immediate previous term

Answer 29

but since u have the general formula given, you can just get it from it. \[a(r)^{n-1}\]

Answer 30

that is the general form of any term of geometric series. where r is common ration

Answer 31

ratio*

Answer 32

okay so r is 1?

Answer 33

No the complete thingy with in the braces ;P

Answer 34

first of all if u keep multiplying by 1, that would mean All the Terms have same value

Answer 35

because 1 raised to any power is 1

Answer 36

@shaeelynn were u able to solve it?

Answer 37

no i still dont really get it

Answer 38

The common ratio r is 1/4 because each new term in the geometric series is generated by multiplying the previous term by 1/4.
You could find T2/T1 or T3/T2 etc. and that will also give you 1/4.

So now you can substitute r = 1/4 and a = 3 to find the limiting sum.

Answer 39

infinite sum of (3 (1/4)^(n-1)) = 3 * infinite sum of (1/4)^(n-1)

this infinite series ( (1/4)^(n-1)) has
first term = (1/4)^0 = 1
common ratio = 1/4

infinite sum = (first term) / (1 - common ratio) = 1 / (1- 1/4)) = 1 / (3/4) = 4/3

your result = 3 * infinite sum above = 3 * 4/3 = 4