I. Limits of Summation

S = a

_{1}+ a

_{1}r + a

_{1}r

^{2}+ ... =

_{¥}

`å`(a

_{1}(r

^{i-1})) =

^{i = 1}

a

_{1}/(1-r) provided that the absolute value of r

`<`1

Using Limit Notation, this sum can be written as:

S =

`¥`

lim

`å`(a

_{1}r

^{i-1})

^{n®¥ i = 1}

= lim a

_{1}(1 - r

^{n})/(1-r)

^{n®¥}

Recall that since r

`<`1 thus making it a fraction, when you take the limit or r

^{n}as x approaches infinity, this would equal zero. So:

= (a

_{1}(1 - 0))/ (1 - r)

= a

_{1}/(1 - r) which you have already seen in this coursework.

Summation Formulas and Properties:

Example 1: Evaluating a Summation:

What is the sum of

i^{2} 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)/(n^{2})^{i = 1}

= (1/n^{2})(2(n(n+1))/2) + 3n) = (1/n^{2})(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/n^{3})(1 + i^{2})=^{i = 1}

= (3/n^{3})(n + (n(n + 1)(2n + 1)/6) =

(3/n^{2} + (3(2n^{2} + 3n + 1)(6n^{2})

= 3/n^{2} + (6n^{2} + 9n + 3)/(6n^{2})

lim 3/n^{2} = 0 and^{n ®¥}

lim (6n^{2} + 9n + 3)/(6n^{2}) = 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)...}