Sequences

Sequences
A sequence of real numbers is a function a? N ? R.
The sequence is denoted by {a_1 ,a_2 ,a_3 ,?,a_n ,?} or { a_n }
For example, the expression {2n} denotes the sequence {2 ,4 ,6 ,?}
The number a_n is called the general term of the sequence { a_n }.
Example 1: Write down the first few terms of each of the following sequences.
1. {(n+1)/n^2 } 2. {2n/(n+3)}
3. {(-1)^(n+1)/2^n } 4. { c_n } where c_n=nth digit of ?
Solution
1. {(n+1)/n^2 }={?2?(n=1),?(( 3 )/( 4 ))?(n=2),?(( 4 )/( 9 ))?(n=3),?(( 5 )/( 16 ))?(n=4),?(( 6 )/( 25 ))?(n=5),?}
2. {2n/(n+3)}={?(( 1 )/( 2 ))?(n=1),?(( 4 )/( 5 ))?(n=2),?1?(n=3),?(( 8 )/( 7 ))?(n=4),?(( 5 )/( 4 ))?(n=5),?}
3. {(-1)^(n+1)/2^n }={?(( 1 )/( 2 ))?(n=1),?((-1 )/( 4 ))?(n=2),?(( 1 )/( 8 ))?(n=3),?((-1 )/( 16 ))?(n=4),?(( 1 )/( 32 ))?(n=5),?}
4. { c_n }={3,1,4,1,5,?}

convergent and divergent sequence:
If exists and is finite we say that the sequence is convergent. If doesn’t exist or is infinite we say the sequence diverges. Note that sometimes we will say the sequence diverges to ? if and if we will sometimes say that the sequence diverges to -? .


Example 2: Determine whether the sequence converges or diverges.
1. {(n^2+1)/(n+1)^2 } 2. {(n+17)/?(2n^2+3n)} 3. {e^n/n^3 }
4.{ ?(n+4) -?(n ) } 5.{ ?(n^2+5n) -n }
Solution
1. lim?(n??)??(n^2+1)/(n+1)^2 ?=lim?(n??)??2n/2(n+1) ?
=lim?(n??)??( 2 )/2?=1 converges to 1
2. lim?(n??)??(n+17)/?(2n^2+3n)?=lim?(n??)??(1+17/n)/?(2+( 3 )/n)?
=( 1 )/?( 2 ) converges to ( 1 )/?( 2 )
3. lim?(n??)??e^n/n^3 ?=lim?(n??)??e^n/?3n?^2 ?=lim?(n??)??e^n/6n?
=lim?(n??)??e^n/6?=? diverges
4. lim?(n??)?(?(n+4) -?(n ))=lim?(n??)?(?(n+4) -?(n ))×((?(n+4)+?(n )))/((?(n+4)+?(n )) )
=lim?(n??)??(n+4-n)/((?(n+4)+?(n )) )?=0 converges to 0
5. lim?(n??)??(?(n^2+5n) -n)×(?(n^2+5n)+n)/(?(n^2+5n)+n)?=lim?(n??)??(n^2+5n-n^2)/(?(n^2+5n)+n)?
=lim?(n??)??5n/(?(n^2+5n)+n)?=lim?(n??)??5/(?(1+5/( n ))+1)?
=5/( 2 ) converges to 5/( 2 )

Limits that arise frequently
1. lim?(n??)??ln?n/n?= 0 2. lim?(n??)??(n&n)=1
3. lim?(n??) x^(1/n)=1 ; (x>0) 4.lim?(n??) x^n=0 ? |x|<1
5.lim?(n??)??(1+( x )/n)^n=e^x ? 6. lim?(n??) x^n/n!=0
Calculation of the limits
1.lim?(n??)??ln?n/n?=lim?(n??)??(( 1 )/n)/1?=0
2.Let a=lim?(n??)??(n&n) then ln?a=ln?lim?(n??)??(n&n)
ln?a=ln?lim?(n??)??(n)^(1?n) ?
ln?a=lim?(n??)?ln??(n)^(1?n) ?
ln?a=lim?(n??)??ln?n/n?= 0
a=e^0=1
So lim?(n??)??(n&n)=1
5. Let a=lim?(n??)??(1+( x )/n)^n ? then ln?a=lim?(n??)?ln??(1+( x )/n)^n ?
ln?a=lim?(n??)??n ln?(1+( x )/n) ?=lim?(n??)??ln?(1+( x )/n)/(( 1 )/n)?
ln?a=lim?(n??) (1/((1+x/n) ).(-x)/n^2 )/((-1)/n^2 )=x
a=e^x
So lim?(n??)??(1+( x )/n)^n ?=e^x

Example 3: Determine whether the sequence converges or diverges.
1.{(1+ln?n)/n} 2. {1/(0.6)^n } 3. {((? 2?^n )/( n ))^(1?n) }
4. {?(n&n^2+n )} 5. {ln??((n-2)/n)^n ? } 6. {n!/?10?^n }
Solution
1. lim?(n??)???1+ln??n/n?=lim?(n??) ( 1 )/n+ lim?(n??)??ln?n/n?= 0 converges to 0
2. lim?(n??)??1/(0.6)^n ?=1/(lim?(n??) (0.6)^n )=( 1 )/0=? diverges
3. lim?(n??) ((? 2?^n )/( n ))^(1?n)=lim?(n??) (2 )/( ?(n&n) )=2 converges to 2
4. lim?(n??)??(n&n^2+n)=lim?(n??)??(n&n(n+1) )=lim?(n??)??(n&n)×lim?(n??)??(n&n+1)=1 converges to1
5. lim?(n??)??ln??((n-2)/n)^n ?=ln??lim?(n??) (1+(-2)/n)^n ? ?=ln??e^(-2) ?=-2 converges to -2
6. lim?(n??) n!/?10?^n =1/(lim?(n??) ?10?^n/n!)=( 1 )/0=? diverges


Determine whether the sequence converges or diverges.
1.{(3+ln??n^n ?)/n^2 } 2. {(( 3 )/n)^(1/n) } 3. {((n+5)/n)^n }
4. {?(n&4^n n)} 5. {(n^5+2n)/(3n^4+n^2 )} 6. {8^(n+1)/n!}
7.{(e^n+n^2)/(e^n-2n^2 )} 8. {?(n(n+2) ) -n} 9. {(3^n/(2n+1))^(1?n) }
10. {(3n^2-ln?n)/(n^2+3n^(3?2) )} 11. {((n-3)/n)^n }