nth Term Divergence Test (nTDT)

nth Term Divergence Test (nTDT) Consider a series ∞ X an . If lim an 6= 0 then the series Diverges. n=1 n→∞ USED: When you suspect the terms of the given series do not approach zero. This is a quick and straightforward test, assuming the limit of the terms is a manageable computation. Understand how the behavior of the terms can determine divergence of the series. Generally, this test is helpful when the series seems a bit “oddball” in form or is not a more natural candidate for another convergence test. Sometimes the conditional statement can be confusing. Check the if-then ordering carefully. NOTE: This test is Inconclusive if the terms DO approach zero. If lim an = 0, then you have n→∞ more work to do with a different Convergence test. Simply put, if the terms do approach 0 then you know nothing except that there is posssible convergence. Remember, the Harmonic Series ∞ X 1 1 is a divergent series even though the terms shrink to zero as n → ∞. n n n=1 WARNING: Do not create an nth Term Convergence Test. There is no such result or test. Why? IMPORTANT: There is only a Divergence conclusion here, for nTDT. WARNING: Do not declare Convergence from a Divergence test. APPROACH: • Given a series ∞ X an , step aside and examine the terms of the series. Study lim an . n→∞ n=1 • If you see that the terms will approach zero as n approaches infinity, then do not waste your time proving that the terms approach zero. This test will not help you. Move onto another, more appropriate, convergence test. • If you suspect that the terms do not go to zero as n approaches infinity, then carefully compute lim an . Show the actual limit answer, AND write that lim an 6= 0 does not equal n→∞ n→∞ zero. That is the condition of the test you need to write clearly. Use all of your limit training for sequences. If you need L’H Rule, then make sure that you switch to the related function having x as the variable. • Think about this Divergence test in the following way... If lim an = L 6= 0, then it says n→∞ that the sequence terms {an }∞ n=1 Converge to the limit L. However, because the terms do not converge to 0, then the original series Diverges. FINAL THINK:   as n → ∞, then the Series Diverges −→0, •If the terms an        •If the terms an −→ 0, then STOP nTDT, result is INCONCLUSIVE    ,→ need a DIFFERENT TEST EXAMPLES: Determine and state whether each of the following series Converges or Diverges. Name any Convergence test(s) that you use, and justify all of your work. ! √ 2 ∞ X 3 n +1 √ 1. arctan Diverges by nth term Divergence Test since 2+ 3 n n=1 !
! √ √ 2 1 3 n +1 ( 3 n2 + 1) 2 √ √ = arctan lim · n1
lim arctan n→∞ (n2 + n) n→∞ n2 + 3 n2   0 √ 7  √ 3 + n12 π   = arctan( 3) = = arctan  lim 6= 0 0 n→∞ 3 7  1 + 1 3 n2  2. ∞ X n=1   1 n · arctan n Diverges by nTDT since     1 1 ∞·0 lim an = lim n · arctan = lim x · arctan n→∞ n→∞ x→∞ n x   1
1  0  2 −
2
x 1 0 1 arctan 1+ = 1 x x = lim = lim L’H lim 0= 1 6= 0 1 x→∞ x→∞ x→∞ 1
1 −
2 7 1 + x2 x
x  n ∞  X n 3. n+5 Diverges by the nth term Divergence Test since n=1  lim an = lim n→∞ n→∞ n n+5 n  = lim x→∞ x x+5 x(1∞ ) „ ln = lim e x «x x+5 x→∞ ∞ ∞ x ln x+5 x   (∞·0)  lim x x 1 x→∞ lim x ln lim ln x→∞ x→∞ x + 5 x + 5 x =e =e =e       x+5 ( x + 5)(1) −  x(1) x+5 5 x (x + 5)2 x (x + 5)2 lim lim 1 1 x→∞ x→∞ − 2 − 2 L’H x x = e =e !    ∞  x+ 5 5  −5x ∞ lim (−x
2 ) lim x→∞ 1 x L’H −5  (x + 5)
2 =e = ex→∞ x + 5 = e 1 = 5 6= 0 e  4. ∞ X n=1 8 Diverges by the nth term Divergence Test since lim 8 = 8 6= 0 n→∞