Convergence Tests Academic Resource Center
Series Given a sequence {a 0, a, a 2,, a n } The sum of the series, S n = A series is convergent if, as n gets larger and larger, S n goes to some finite number. If S n does not converge, and S n goes to, then the series is said to be divergent n k a k
Geometric and P-Series The two series that are the easiest to test are geometric series and p-series. Geometric is generally in the form P-series is generally in the form k ar k n n p
Geometric Series A geometric series is a series in which there is a constant ratio between successive terms +2 + 4 + 8 + each successive term is the previous term multiplied by 2 each successive term is the previous term squared. 2 4 6 256...
Geometric Series S n = = ar ar +ar^2 + ar^3 + +ar^k k k As a result, if r <, the geometric series will converge to, and if r the series will diverge. a r
P-Series Given a series n n p This series is said to be convergent if p>, And divergent if p
Geometric and P-Series Examples 3 n 4 3 3 n 4 n n So S= 3/(-3/4) = 2 n This series is geometric with a=3 and r =3/4. Since r<, this series will converge. The Sum of the series, S a S r n n 3 Here, p=3, so p>. Therefore our series will converge n n n n 2 Here, p=/2, so p<. Therefore our series will diverge
Convergence Tests Divergence test Comparison Test Limit Comparison Test Ratio Test Root Test Integral Test Alternating Series Test
Divergence Test Say you have some series The easiest way to see if a series diverges is this test Evaluate L= Lim If L 0, the series diverges If L=0, then this test is inconclusive n a n n 0 a n
Divergence Test Example n n 2 5 n 2 4 Let s look at the limit of the series Lim n n 2 5 n 2 4 Lim n n 2 5 n 2 5 0 Therefore, this series is divergent n Lim n n 2 n 0 2 The limit here is equal to zero, so this test is inconclusive. However, we should see that this a p-series with p>, therefore this will converge.
Comparison Test Often easiest to compare geometric and p series. Let and be series with non-negative terms. If a k for k, as k gets big, then b k If converges, then converges If diverges, then diverges b k a k b k a k b k a k
Comparison Test Example n 3 n This is very similar to And since n 3 n Test to see if this series converges using the comparison test n n 3 n 3 n which is a geometric series so it will converge our original series will also converge
Limit Comparison Test Let and be series with non-negative terms. a k Evaluate Lim If lim=l, some finite number, a then both k either converge or diverge. b k and and are generally geometric series or p-series, so seeing whether these series are convergent is fast. b k k b k a k a k b k
Limit Comparison Test Example n 9 n 3 0 n Compare it with And since n Determine whether this series converges or not n 9 n 0 n n 9 0 n n 9 0 so Lim n Is a geometric series with r<, this series converges, therefore so does our original series 9 n 3 0 n 0 n 9 0
Ratio Test Let a k be a series with non-negative terms. Evaluate L= Lim If L <, then If L >, then converges diverges If L =, then this test is inconclusive k a k a k a k a k
Ratio Test Example Test for convergence n n 3 3 n n Look at the limit of a n a n Lim n () n (n ) 3 3 n () n n 3 3 n Lim n (n ) 3 3 n 3 n n 3 3 Lim ( n n n ) 3 3 Lim ( n n ) 3 3 Since L<, this series will converge based on the ratio test
Root Test Let a n Useful if Evaluate L= Lim If L <, If L >, be a series with non-negative terms. involves nth powers a n is convergent n is divergent If L =, then this test is inconclusive a n a n (a n ) n
Root Test Example Test for convergence n ( 4 n 5 5 n 6 ) n Lim (( 4 n 5 n 5 n 6 ) n ) Lets evaluate the limit, L =Lim n Lim n By the root test, since L<, our series will converge. n (a n ) 4 n 5 5 n 6 4 5 n
Integral Test Given the series, let = f(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 general, however, k 0 a k 0 f ( x )dx 0 a k f ( x )dx a k f ( x k )dx
Integral Test Example Test for convergence n (2 n ) 3 So let f ( x ) (2 x ) 3 Since x>0, f(x) is continuous and positive. f (x) is negative so we know f(x) is decreasing. Now let s look at the integral (2 x ) 3 Lim t ( dx 2 2 Lim [ t (2 t ) 2 (3) ) 2 9 Note: Series will most likely not converge to /9, but it will converge nonetheless. ] (2 x ) 2 Since the integral converged to a finite number, our original series will also converge t
Alternating Series Test Given a series, where is positive for all k IF for all k, and Lim = 0 a k a k () k a k Then the series is convergent a k k a k
Alternating Series Test Example Test for convergence n n n 2 n 3 4 Check: Is this series decrease- yes Is the Lim=0? Lim n n 2 Therefore, n 3 4 0 n Yes n n 2 n 3 4, is convergent.
More Examples. n cos n n 3 4 2. n n n 2 n 2 n 6 3. n 0 n (n )4 2 n 4. n 2 n ln n
Answers. 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