Precalculus 12.5 The Area Problem

12.5 The Area Problem

I. Limits of Summation

S = a₁ + a₁r + a₁r² + ... =

_¥
å (a₁(r^i-1)) =
^{i = 1}

a₁/(1-r) provided that the absolute value of r < 1

Using Limit Notation, this sum can be written as:

S =
¥
lim å (a₁ r^i-1)
^{n®¥ i = 1}

= lim a₁ (1 - rⁿ)/(1-r)
^n®¥

Recall that since r < 1 thus making it a fraction, when you take the limit or rⁿ as x approaches infinity, this would equal zero. So:

= (a₁ (1 - 0))/ (1 - r)

= a₁/(1 - r) which you have already seen in this coursework.

Summation Formulas and Properties:

Example 1: Evaluating a Summation:

What is the sum of
i² when i = 1 to i = 30?

=(30)(30 + 1)(2(30) + 1)/6 = (30)(31)(61)/6 = 56730/6 = 9455

Example 2: Evaluating a Summation:

What is the sum of:
(2i + 1) when i = 1 to i = 50?

You have to break this down into parts:

2( å i ) from i = 1 to i = 50 AND
å(1) from i = 1 to i = 50 is equal to

2(50(50+1)/2) + 1(50) = (50)(51) + 50 = 2550 + 50 = 2600

II. Finding the Limit of a Summation

_n
å (2i + 3)/(n²)
^{i = 1}

= (1/n²)(2(n(n+1))/2) + 3n) = (1/n²)(n(n+1) + 3n)
= (1/n)(n + 1 + 3) = (n + 4)/n

which in essence is infinity over infinity so
Therefore the Summation of
lim S_n =
^n®¥

lim (n + 4)/n = 1/1 = 1 so
^n®¥

lim S_n = 1
^n®¥

You can check this by using a table of values:

(n, S(n)) = {(1, 5), (10, 1.4), (100, 1.04), (1000, 1.004), (10000, 1.0004)...}
So you can see the values of S(n) approach 1.

Example 3: Finding the Limit of a Summation:

_n
å (3/n³)(1 + i²)=
^{i = 1}

= (3/n³)(n + (n(n + 1)(2n + 1)/6) =
(3/n² + (3(2n² + 3n + 1)(6n²)
= 3/n² + (6n² + 9n + 3)/(6n²)

lim 3/n² = 0 and
^{n ®¥}

lim (6n² + 9n + 3)/(6n²) = 6/6 = 1
^{n ®¥}

So:

lim S_n = 1
^{n ®¥}

Again, you can check this using a table of values:
(n, S(n)) = {(1, 6), (10, 1.185), (100, 1.0154), (1000, 1.0015), (10000, 1.0002)...}