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The nth term divergence test is a method used to determine the convergence or divergence of a series by examining the limit of the terms as n approaches infinity. If the limit is not equal to zero, then the series diverges. This test is particularly useful when the terms do not appear to approach zero, or when the series is an 'oddball' form.
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nth^ Term Divergence Test
Consider a series
n=
an. If (^) nlim→∞ an 6 = 0 then the series diverges.
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.
NOTE: This test is inconclusive if the terms do approach zero. If (^) nlim→∞ an = 0, then you have 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 ∑^ ∞
n=
n is a divergent series even though the terms
n shrink to zero as^ n^ → ∞.
WARNING: Do not create an nth^ term convergence test. There is no such result or test. Why?
WARNING: Do not declare convergence from a divergence test.
APPROACH:
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.
n=
arctan
3 n^2 + 1 n^2 +
Diverges by nth^ term Divergence Test since
nlim→∞ arctan
3 n^2 + 1 n^2 +
= arctan
(^) nlim→∞
3 n^2 + 1) (n^2 +
n) ·
n^2
n^2
= arctan
(^) lim n→∞
3 + (^) n^12 1 + 1 n 32
(^) = arctan(√3) = π 3 6 = 0
n=
n · arctan
n
Diverges by the nth^ term Divergence Test since
nlim→∞ an^ = lim n→∞ n^ ·^ arctan
n
∞· = (^0) lim x→∞ x^ ·^ arctan
x
= lim x→∞
arctan
x
x
L’H lim x→∞
x
x^2
− (^) x^12
= lim x→∞
1 + (^) x^12
n=
n n + 5
)n Diverges by the nth^ term Divergence Test since
nlim→∞ an^ = lim n→∞
n n + 5
)n = lim x→∞
x x + 5
)x(^1 ∞) = lim x→∞ e
ln
„ (^) x x + 5
«x
= e x^ lim→∞ ln
x x + 5
)x]
= e x^ lim→∞ x^ ln
x x + 5
= e
x^ lim→∞
ln
x x + 5
x
∞∞
L’H = e
x^ lim→∞
x + 5 x
(x + 5)(1) − x(1) (x + 5)^2
x^2 = e
x^ lim→∞
x + 5 x
(x + 5)^2
x^2
= e x^ lim→∞
x + 5 x
(x + 5)^2
(−x^2 ) = e x^ lim→∞
− 5 x x + 5
L’H = e − (^15) =^1 e^5 6 = 0