Quick viewing(Text Mode)

Convergence Tests Academic Resource Center Series

Convergence Tests Academic Resource Center

• Given a {a0, a1, a2,…, an }

• The sum of the series, Sn= n a  k k 1 • A series is convergent if, as n gets larger and larger, Sn goes to some finite number. • If S does not converge, and S goes to ∞, then the series is n  n said to be divergent Geometric and P-Series

• The two series that are the easiest to test are and p-series.

• Geometric is generally in the form   ar k • P-series is generally in the form k 1

 1   p n 1 n

 Geometric Series

• A geometric series is a series in which there is a constant ratio between successive terms

• 1 +2 + 4 + 8 + … each successive term is the previous term multiplied by 2

• each successive term is the previous term squared.

1 1 1 1     ... 2 4 16 256

 Geometric Series

 • k Sn=  =ar ar +ar^2 + ar^3 + … +ar^k k 1 • As a result, if |r|<1, the geometric series will converge to , and if |r| 1 the series will diverge. 

a 

1  r

  P-Series

 1 • Given a series  n p n 1 • This series is said to be convergent if p>1,

 • And divergent if p 1

 Geometric and P-Series Examples

 n  n 1 This series is 3  3  The Sum of the series, S    3  geometric with a=3 4 n 1 4  and r =3/4. Since a n 1 n 1 r<1, this series will S  converge. So S= 3/(1-3/4) = 12 1  r

 1 Here, p=3, so p>1. Therefore our series   3 will converge n 1 n 

1   1 1 2 Here, p=1/2, so p<1. Therefore our series will      diverge n n  n 1 n 1



test • Comparison Test • Comparison Test • Test • Test Divergence Test

 • Say you have some series  a n • The easiest way to see if a series divergesn  0 is this test • Evaluate L= Lim • If L 0, the series diverges • If L=0, then this test is inconclusive a n n  

 



 Divergence Test Example

 n 2 Let’s look at the limit  2 n 1 5 n  4 of the series

n 2 n 2 1 Lim 2  Lim 2   0 Therefore, this series is divergent n   5 n  4 n   5 n 5 

 1   n 2 The limit here is equal to zero, so this test is inconclusive. n 1 However, we should see that 1 this a p-series with p>1, therefore this will converge. Lim 2  0 n   n

 Comparison Test

• Often easiest to compare geometric and p series.

• Let and be series with non-negative terms. • If for k, as k gets big, then  ak  bk • If converges, then converges • If diverges, then diverges b  a  k  k

 Comparison Test Example

 1 Test to see if this series converges  n using the comparison test n 1 3  1  1 which is a geometric series so This is very similar to  n it will converge n 1 3   1  1 And since    our original series will also converge 3 n 3 n  1  n 1 n 1



• Let and be series with non-negative terms.  ak  bk • Evaluate Lim • If lim=L, some finite number,a then both and either converge or diverge.k   k   b • and are generallyk geometric series or p-series, so

seeing whether these series are convergent is fast.

 Limit Comparison Test Example  9 n Determine whether this series  n n n 1 3  10 converges or not 9   n 9 n  9  3  10 n Lim n  1  0 Compare it with     so n   n    9  n 1 10 n 1 10   10   n   9  And since    Is a geometric series with r<1, this series converges,   n 1 10 therefore so does our original series  

 Ratio Test

• Let be a series with non-negative terms.  ak

• Evaluate L= Lim a • If L < 1, then convergesk 1  a • If L > 1, then kdiverges  k • If L = 1, then this test is inconclusive 

 Ratio Test Example

Test for convergence  3 a n n Look at the limit of n 1 1    n a n n 1 3

(1) n 1 (n  1) 3 3 n 3 n 1 (n  1) 3 Lim n 3  Lim n 1  3 n   (1) n n   3 n Since L<1, this series  3 n will converge based on the ratio test 1 n  1 1 1 1  Lim ( ) 3  Lim (1  ) 3   1 3 n   n 3 n   n 3

 Root Test

• Let be a series with non-negative terms.  an • Useful if involves nth powers

a n • Evaluate L= Lim 1 • If L < 1, is convergent n  (a n ) • If L > 1, is divergentn   a • If L = 1, then thisn test is inconclusive    an  



Root Test Example

Test for convergence

 1 4 n  5 n n Lets evaluate the limit, L =Lim (a n )  ( ) n   n 1 5 n  6   4 n  5 1 4 n  5 4 Lim (( ) n ) n  Lim   1 n   5 n  6 n   5 n  6 5

 By the root test, since L<1, our series will converge.

 Integral Test

• Given the series , let =a f(k)  k a k • f must be continuous, positive, and decreasing for x > 0 • will converge only if converges. • If diverges, then the series will also diverge.    • In ageneral, however,  k  f ( x)dx k  0 0

    a  f ( x )dx  k 1 k 1

 Integral Test Example

Test for convergence Since x>0, f(x) is  1 1 continuous and So let positive.  3 f ( x )  (2 n  1) (2 x  1) 3 f’(x) is negative so n 1 we know f(x) is decreasing.

Now let’s look  1 1 1 t  3 dx    2 Lim [ 2 ] at the integral 1 (2 x  1) 2 t   (2 x  1) 1   1 1 1   Lim ( 2  2 )  t   (2 t  1) (3) 9

Since the integral converged to a finite number, our original series will also converge

Note: Series will most likely not converge to 1/9, but it will converge nonetheless.

• Given a series (1) k a , where is positive for all k  k a k • IF • for all k, and

• Lim = 0 

a  a  Then kthe1 seriesk is convergent

k  

  Alternating Series Test Example Test for convergence  2 n 1 n Check:  1 3 Is this series decrease- yes n 1 n  4 Is the Lim=0?

n 2 Lim 3  0 Yes n   n  4 

Therefore, , is convergent.  More Examples

 cos n  1  n  n 2 1. 2.  3 4  2 6 n 1 n n 1 1  n  n

  n  10  1 4.  3.  2 n 1 n  2 n ln n n 1 (n  1)4

  Answers

• 1. By Alternating series test, series will converge • 2. By the comparison test, series will diverge • 3. By the ratio test, series will converge • 4. By the integral test, series will diverge