Exam 3

Question 1

∑ (n=2, ∞) (1 + C)^-n = 2

Answer 1

Geometric Series
∑ (n=2, ∞) ( 1 / 1+C) ^n
s = (1-r^n) / (1-r)
s = ∑ (i=1, n) a(i)
= ∑ (i=1, n) a(1) * r^(i-1)
∑ (n=2, ∞) ( 1 / 1+C )^n + 1 + 1 / 1+C = ∑ (n=0, ∞) ( 1 / 1+C )^n
S + 1 + 1/1+C = 1/ ( 1 - (1/1+C) )
Multiply both sides by (1+C)
3 + 3C +1 = (1+C) / ( 1 - (1/1+C) ) = C
4+3C = (1 +C) / (1- (1/1+C)) * (1+C) / (1+C) = 1 + 2c + c^2 / 1 + c -1
4 +3C = (1 + 2c + c^2) / c
4c + 3c^2 = 1 + 2c + c^2
2c^2 + 2c -1 = 0
apply quadratic formula
C = ( -2 +- sqrt ( 4 - 42-1) ) / 4
= (-2 +- sqrt (12) ) / 4
= 0.366, -1.366
1/1+C = 0.732, - 2.7322
Therefore , c = 0.366 = -1/2 + sqrt(3)/2

Question 2

1 + 2/7 + 2^2/7^2 + 2^3/7^3

Answer 2

Geometric Series
r = 2/7 less than one therefore converges
s = 1 + r + r^2 + … r^n
= 1 / 1 - (2/7) = 1/ (5/7)
= 7/5

Question 3

1/16 + 3/64 + 9/256 + 3/1024 + …

Answer 3

Geometric Series
S = 1/16 ( 1 + 3/4 + 3^2/4^2 + 3^3/4^3 + … )
S = 1/16 ( 1 + r + r^2 + r^3 + … )
r = 3/4 less than one therefore converges
1/16 * 1 / 1 - (3/4)
1/16 * 1/ (1/4)
1/16 *4
= 1/4

Question 4

∑ (m=2, ∞) 5 / 2^m

Answer 4

Geometric Series
S = 5 ∑ (m=2, ∞) 1 / 2^m
S = 5 ∑ (m=2, ∞) (1/2)^m
∑ (m=2, ∞) = 1/4 + 1/8 + 1/16
= 1/4 ( 1 + 1/2 + 1/4 + 1/8 + …)
= 1/4 ∑ (m=0, ∞) (1/2)^2
S = 5/4 ∑ (m=0, ∞) (1/2)^m
r=1/2 less than one therefore converges
= 5/4 * 1 / 1 - (1/2)
= 5/4 * 2
=5/2

Question 5

∑ (k=3, ∞) 3 * 4^k / 7^k

Answer 5

Geometric Series
S = 3 ∑ (k=3, ∞) (4/7)^k
r = 4/7 , which is less than one therefore converges
S = 3 * (4/7)^3 ∑ (k=0 ∞) (4/7)^k
1/ 1 - (4/7) = 1/ (3/7) = 7/3
S = 3 * (4/3)^3 * 7/3 = 4^3 / 7^2
64/49

Question 6

∑ (k=1, ∞) (-e)^-k

Answer 6

Geometric Series
S = ∑ (k=1, ∞) ( 1 / -e ) ^k
r = (1/-e) , which is less than one therefore converges
S = 1 / -e * ∑ (k=0, ∞) ( 1 / -e)
= 1/-e * 1 / 1 + (1/e)
= - 1 / (e+1)

Question 7

∑ (n=1, ∞) ln ( 1 + 1/n )

Answer 7

Telescoping Series
∑ (n=1, ∞) ln ( n +1 / n ) = S
1 + 1/ n = n / n + 1 / n = ( n+1 / n )
∑ (n=1, ∞) [ ln (n+1) - ln (n) ]
S = (ln2 - ln1) + (ln3 - ln2) + (ln4 - ln3) + (ln5 -ln4) + …
= infinity , therefore diverges

Question 8

∑ (n=1, ∞) 1 / (n(n+1))

Answer 8

Telescoping Series
Partial Fraction Expansion
1 / n(n+1) = A / n + B / (n+1)
1 = A (n+1) + B (n)
if n = 0 , A =1
if n = -1 , B= -1
S = ∑ (n=1, ∞) ( 1/n - 1/(n+1) ) = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + …
= 1 - infinitely small sum
Therefore = 1

Question 9

∑ (n=2, ∞) 1 / n* ln(n)

Answer 9

Integral Test
∫ (1, ∞) 1 / xlnx *dx
u = lnx
du = 1/x dx
x =1 -> u = 0
∫ (0, ∞) u^-1 du
ln(u) ] (0, ∞)
Therefore both cases diverge and the whole series diverges

Question 10

∑ (n=1, ∞) n^2 -1 / 3n^4 +1

Answer 10

Comparison Test
-> n^2 / 3n^4 -> 1/3 * 1/n^2
1/n^2 > 1 / 3n^2 = n^2 / 3n^4 > n^2-1 / 3n^4 > n^2 -1 / 3n^4 +1
1/n^2 converges , therefore the original series converges by comparison theorem

Question 11

∑ (n=1, ∞) 1 / n(n+1)

Answer 11

Comparison Test
∑ (n=1, ∞) 1 / n (n+1)
=1 , previous telescoping series done in class
Therefore converges

Question 12

∑ (n=1, ∞) 1 / (n+1)(n+1)

Answer 12

Comparison Test
-> 1/ n(n+1) > 1 / (n+1) (n+1)
Therefore, the original series converges by the comparison theorem

Question 13

1 / 1 + 2 + 3 + … + n

Answer 13

Comparison Test
First make a summation for the given series
1 + 2 + 3 + … + n = S
n + n-1 + n-2 + … + 1 = S
n(n+1) = 2 S
S = n(n+1) / 2
∑ (n=1, ∞) 1 / n(n+1) /2
2 ∑ (n=1, ∞) 1 / n(n+1)
-∑ (n=1, ∞) 1 / n(n+1) = 1
=2 * 1
=2
Therefore the original series converges

Question 14

∑ (n=1, ∞) n^2 / n^3 +1

Answer 14

Limit Comparison Test
1 / n = 1 / n * n^2 / n^2 = n^2 / n^3 > n^2 / n^3 +1
1 / n / n^2 / n^3 +1
lim n -> ∞ n^3 +1 / n^3 =1
Therefore the original series diverges by limit comparison test, 1/n diverges

Question 15

∑ (n=1, ∞) sin(1/n)

Answer 15

Limit Comparison Test
A(n) = sin(1/n)
b(n) = 1/n
lim n -> ∞ sin(1/n) / 1/n = 1
Small angle Approximation : lim theta -> 0 (sin(theta) / theta)) =1
Therefore ∑ (n=1, ∞) sin(1/n) diverges by the limit comparison test

Question 16

∑ (n=1, ∞) 1/n

Answer 16

P-Test
Diverges
p=1

Question 17

∑ (n=1, ∞) (-1)^n-1 * 1/n

Answer 17

Alternating Series
Converges
Satisfies both alternating series conditions

Question 18

∑ (n=1, ∞) (-1)^n-1 / 2n+1

Answer 18

Alternating Series
b(n) = 1/ 2n+1
∑ (n=1, ∞) (-1)^n-1 * 1/2n+1
converges, satisfies both alternating conditions

Question 19

∑ (n=1, ∞) (-1)^n cos( pi/n )

Answer 19

Alternating Series
Diverges

Question 20

∑ (n=1, ∞) (-3)^n / n^3

Answer 20

Alternating Series
-3^n = (-1 * 3)^n = -1^n * 3^n
∑ (n=1, ∞) (-1)^n * 3^n / n^3
b(n) = 3^n / n^3 , lim n -> ∞ 3^n / n^3
Indeterminite
lim n -> ∞ ( ∞/∞ )
L’Hopitals Rule
d/dx [ 3^n / n^3]
d/dx [3^n * ln 3 / 3 n^2]
lim n-> ∞ 3^n (ln3)^3 / 321 = ∞

Question 21

∑ (n=1, ∞) n! / 100^n

Answer 21

Ratio Test
A(n) = n! / 100^n
A(n+1) = (n+1)! / 100^(n+1)
lim n -> ∞ | (n+1)! / 100^(n+1) / n! / 100^n |
lim n -> ∞ | (n+1)! / n! * 100^n / 100^(n+1) |
lim n -> ∞ | (n+1) / 100 |
= ∞ , therefore the series diverges

Question 22

∑ (n=1, ∞) (-1)^(n-1) * sqrt(n) / n+1

Answer 22

First Limit Comparison Test
lim n -> ∞ 1/sqrt(n) / sqrt(n)/n+1
=lim n -> ∞ n+1 / n
divergent (p- test), both are divergent
Ratio Test
lim n -> ∞ | sqrt(n+1) / (n+1)+1 / sqrt(n) / (n+1)
lim n -> ∞ | (n+1) / n+2 * sqrt (n+1) / sqrt (n)
= 1 * 1
= 1
Therefore the test is inconclusive

Question 23

∑ (n=1, ∞) sin(4n) / 4^n

Answer 23

Ratio Test
lim n -> ∞ | sin(4(n+1)) / 4^(n+1) / sin(4n) / 4^n |
lim n -> ∞ | sin (4n+1) / sin(4n) * 4^n / 4^(n+1)
| 1 * 1/4 |
=1/4, therefore Absolutely Convergent

Question 24

∑ (n=1, ∞) sqrt (n) * x^n