Infinite Series

Infinite Series
An infinite series is given by the terms of an infinite sequence, added together.
For example, we could take the infinite sequence
{1/2^n }={( 1 )/2 ,( 1 )/4 ,( 1 )/8 ,?}
Then the corresponding example of an infinite series would be given by all of these terms added together
( 1 )/2+( 1 )/4+ ( 1 )/8+?
So we would have ?_(n=1)^??1/2^n =( 1 )/2+( 1 )/4+ ( 1 )/8+?
If ??a_n is a positive series,then either ??a_n converges to a positive number,
or??a_n diverges to infinity.
Divergence Test:
If lim?(n??)??a_n ??0 then ??a_n diverges.
Do not misuse this test. This test only says that a series is guaranteed to diverge if the series terms don’t go to zero in the limit. If the series terms do happen to go to zero the series may or may not converge!
Again, recall the following two series,
?_(n=1)^??( 1 )/n diverges
?_(n=1)^??( 1 )/n^2 converges
One of the more common mistakes that students make when they first get into series is to assume that if then will converges. There is just no way to guarantee this so be careful.
Let’s take a quick look at an example of how this test can be used.
Example 1: Determine if the following series is convergent or divergent.
1. ?_(n=1)^??(n+2)/n 2. ?_(n=0)^??(4n^2-n^3)/(5+2n^3 )
Solution
1. lim?(n??)??(n+2)/n?=1?0
The limit of the series terms isn’t zero and so by the Divergence Test the series diverges.
2. lim?(n??)??(4n^2-n^3)/(5+2n^3 )?=lim?(n??)??(8n-3n^2)/(6n^2 )?=lim?(n??)??(8-6n)/12n?=lim?(n??)??(-6)/12?=-( 1 )/( 2 )?0
By the Divergence Test the series diverges.

Special Series
Geometric Series
A geometric series is any series that can be written in the form,
?_(n=1)^???ar^(n-1) ?=a+ar+ar^2+ar^3+?+ar^n+?
1. If |r|<1 then the geometric series converges to the sum s_n=a/(1-r)
2. If |r|?1 then the geometric series diverges to ? .
Example 2: Determine if the following series converge or diverge. If they converge give the value of the series.
1.?_(n=1)^??( 1 )/2^n 2.?_(n=1)^??(3/( 2 ))^n 3.?_(n=1)^??9^(1-n) 2^(n+1)
Solution
1.The series ?_(n=1)^??( 1 )/2^n converges because r=1/( 2 )<1
a=1/( 2 ),the sum is s_n=(1?2)/(1-(1?2) )=1.
2.The series ?_(n=1)^??(3/( 2 ))^n diverges to ? because r=3/( 2 )>1
3.The series ?_(n=1)^??9^(1-n) 2^(n+1)=?_(n=1)^??( 2^(n+1) )/9^(n-1) =4+( 2^3)/( 9 )+( 2^4)/( 9^2 )+( 2^5)/( 9^3 )+?
converges because r=2/( 9 )<1
a=4,the sum is s_n=4/(1-(2?9) )=4×( 9 )/( 7 )=( 36 )/( 7 )

The p-series
If p is a real constant,the series?_(n=1)^??( 1 )/n^p =( 1 )/1^p +( 1 )/2^p +( 1 )/3^p +?
1.converges if p>1 . 2. diverges if p?1 .
Example 3: Determine whether the series converges or diverges.
1.?_(n=1)^??1/n^2 2. ?_(n=1)^??1/?(n )
1.?_(n=1)^??1/n^2 converges because p=2>1
2. ?_(n=1)^??1/?(n )=?_(n=1)^??1/n^(1?2) diverges to ? because p=1/( 2 )<1










Tests for converges of series
1. Integral Test
Let the function f(x)=a_n (x) be continuous, positive and decreasing then the
series?_(n=1)^??a_n and the integral ?_1^??f(x) dx both converge or both diverge.
Example 4: Determine whether the series converges or diverges.
1. ?_(n=1)^??1/(n^2+1) 2. ?_(n=1)^??n/(n^2+1) 3. ?_(n=1)^??1/(n^2-n-2)
1. ? ?_1^??dx/(x^2+1)=tan^(-1)?x ?|_1^?=?/2-?/4=?/4 converge
2.?_1^??xdx/(x^2+1)=? 1/2 ln?|x^2+1| ?|_1^? =? diverge
3. 1/(x^2-x-2)=A/(x-2)+ B/(x+1) ? x=2 ?A=( 1 )/( 3 ) and x=-1 ?B=-( 1 )/( 3 )
?_1^??dx/(x^2-x-2)=( 1 )/( 3 ) ? ?_1^??(1/(x-2)- 1/(x+1)) dx=( 1 )/( 3 ) ln?|(x-2)/(x+1)| ?|_1^?
=? ( 1 )/( 3 ) ln?|(1-( 2 )/x)/(1+( 1 )/x)| ?|_1^?=( 1 )/( 3 ) ln?2 converge



2. Ratio Test
Let ?_(n=1)^??a_n be a series with non-negative terms and lim?(n??)??a_(n+1)/a_n ?=L .
If 1. L<1 then the series converges
2. L>1 then the series diverges
3. L=1 then this test is inconclusive
Example 5: Determine whether the series converges or diverges.
1. ?_(n=1)^??? n?^3/( 3^n ) 2. ?_(n=1)^??n!/2^n 3. ?_(n=1)^??3/(2n+5) 4.?_(n=1)^??5^n/(n 3^n )
1. lim?(n??)??a_(n+1)/a_n ?=lim?(n??)??? (n+1)?^3/( 3^(n+1) )×( 3^n)/? n?^3 ?=( 1 )/( 3 ) lim?(n??) ((n+1)/n)^3
=( 1 )/( 3 ) lim?(n??) (1+( 1 )/n)^3=( 1 )/( 3 )<1 converges
2. lim?(n??)??a_(n+1)/a_n ?= lim?(n??) (n+1)!/2^(n+1) ×(2^n )/n!
=lim?(n??) ((n+1)×n!)/(2^n×2)×(2^n )/n!=lim?(n??) (n+1)/2=?>1 diverges

3. lim?(n??)??a_(n+1)/a_n ?=lim?(n??)??3/(2(n+1)+5)?×( 2n+5 )/( 3 )
=lim?(n??)??(2n+5)/(2n+7)?=1. The test is inconclusive .
So we try another test ? ?_1^??3dx/(2x+5)=( 3 )/( 2 ) ln?|2x+5| ?|_1^?=? diverges
4. lim?(n??)??a_(n+1)/a_n ?=lim?(n??)??5^(n+1)/((n+1) 3^(n+1) )×(n 3^n)/5^n ?
=lim?(n??)??5n/(3(n+1) )?=( 5 )/( 3 )>1 diverges
3. Root Test
Let ?_(n=1)^??a_n be a series with non-negative terms and lim?(n??)??(n&a_n )=L .
If 1. L<1 then the series converges
2. L>1 then the series diverges
3. L=1 then this test is inconclusive
Example 6: Determine whether the series converges or diverges.
1. ?_(n=1)^??((3n+2)/( 4n+1))^n 2. ?_(n=1)^??(1/( n ))^n 3. ?_(n=1)^??2^n/n^2 4. ?_(n=1)^??((n-1)/( n ))^n
1. lim?(n??)??(n&a_n )=lim?(n??)??(3n+2)/( 4n+1)?=( 3 )/( 4 )<1 converges

2. lim?(n??)??(n&a_n )= lim?(n??) 1/( n )=0<1 converges
3. lim?(n??)??(n&a_n )=lim?(n??)??2/?(n&n^2 )?=2/(lim?(n??) ?(n&n))=( 2 )/( 1 )=2 >1 diverges
4. lim?(n??)??(n&a_n )=lim?(n??) (n-1)/( n )=1.The test is inconclusive .
So we try another test lim?(n??) ((n-1)/( n ))^n=lim?(n??) (1+(-1)/( n ))^n=e^(-1)?0 diverges








Find the sum of the following series:
1. ?_(n=1)^??( 2^n )/? 5?^n 2.?_(n=1)^??( 5^n )/? 3?^2n 3.?_(n=1)^??( 4 )/? 3?^(n+1)
Determine whether the series converges or diverges. Give reasons for your answers.
4. ?_(n=1)^??(n/( 2n-1 ))^n
5. ?_(n=1)^??( 1 )/(? n?^2 ?n)
6. ?_(n=1)^??(n-1)!/( n !)
7. ?_(n=1)^??( ?11?^n )/? 3?^2n
8. ?_(n=1)^???ne^(-n^2 ) ?
9.?_(n=1)^??? e^(-?n)/?n?
10. ?_(n=1)^??((3n+4)/( 2n ))^(-n)
11. ?_(n=1)^??1/( (3n+1)^3 )
12. ?_(n=1)^??(n+3)!/(2^n n!)