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The Ratio Test is a method used to determine the convergence or divergence of a series based on the limit of the ratio of consecutive terms. the theorem, proof, and examples of using the Ratio Test.
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I (^) The ratio test. I (^) Using the ratio test. I (^) Few more examples. I (^) Comment: The root test.
converges or not.
Let {an} be a positive sequence with (^) nlim→∞ an+ an = ρ exists.
(a) If ρ < 1 , the series
an converges. (b) If ρ > 1 , the series
an diverges. (c) If ρ = 1, the test is inconclusive.
an with an appropriate geometric series
r n.
an is non-decreasing. We now show that
an is bounded above. Since (^) nlim→∞ an+ an = ρ < 1, then for any > 0, small enough such that ρ + = r < 1, there exists N large with an+ an < ρ + = r , for n > N.
aN+n aN
aN+ aN
aN+ aN+
aN+n aN+n− 1 6 r n^ ⇒ aN+n 6 aN r n.
∑^ ∞
n=
an =
n=
an +
n=
aN+n 6
n=
an + aN
n=
r n
So
n=
an 6
n=
an + aN 1 − r is bounded. A non-decreasing, bounded above, series converges.
an+ an = ρ > 1, there exists N large, an+ an
1 , for n > N, ⇒ aN < aN+1 < aN+2 < · · ·.
Hence, (^) nlim→∞ an 6 = 0. The series
an diverges.
Case (c): (^) nlim→∞ an+ an = 1. Examples:
∑^ ∞
n=
n
, and (^) nlim→∞
n+
n
) = (^) nlim→∞ n n + 1
= 1, diverges.
n=
n^2 , and (^) nlim→∞
(n+1)^2
n^2
) (^) = (^) nlim→∞^ n
2 (n + 1)^2 = 1, converges.
The test in inconclusive.
Determine whether the series
n=
(n − 1)! (n + 1)^2 converges or not.
Solution: We use the ratio test, since an = (^) ((nn+1)−1)! 2 > 0. Then,
an+ an
n! (n + 2)^2
(n + 1)^2 (n − 1)!
n (n − 1)! (n + 2)^2
(n + 1)^2 (n − 1)!
n(n + 1)^2 (n + 2)^2
an+ an
n^3 + 2n^2 + n n^2 + 4n + 4
n + 2 + (^1) n 1 + (^4) n + (^) n^42
Therefore, (^) nlim→∞ an+ an = (^) nlim→∞ n = ∞.
Since ρ = ∞ > 1, the series diverges. C
Determine whether the series
n=
ln(n) n converges or not.
Solution: We start with the ratio test, since an = ln(n) n
- Then,
an+ an
ln(n + 1) (n + 1)
n ln(n)
n (n + 1)
ln(n + 1) ln(n)
Since ρ = 1, the ratio test is inconclusive.
Direct comparison test: an = ln(n) n
n implies that
∑ (^) ln(n) n
n , which diverges.
Therefore, the series diverges. C
I (^) The ratio test. I (^) Using the ratio test. I (^) Few more examples. I (^) Comment: The root test.
Determine whether the series
n=
ln(7n) n^3 converges or not.
Solution: We start with the ratio test, since an = ln(7n) n^3
an+ an
ln[7(n + 1)] (n + 1)^3
n^3 ln(7n)
( (^) n n + 1
) (^3) ln(7n + 7) ln(7n)
Since ρ = 1, the ratio test is inconclusive. Direct comparison test: Since ln(7n) < 7 n, then
an = ln(7n) n^3
7 n n^3
n^2
. Hence
∑ (^) ln(7n) n^3
n^2
which converges. Therefore, the series converges. C
Let {an} be a positive sequence with (^) nlim→∞^ n
an = ρ exists.
(a) If ρ < 1 , the series
an converges. (b) If ρ > 1 , the series
an diverges. (c) If ρ = 1, the test is inconclusive.
an with an appropriate geometric series
r n.
an is non-decreasing. We now show that
an is bounded above. Since (^) nlim→∞^ n
an = ρ < 1, then for any > 0, small enough such that ρ + = r < 1, there exists N large with √ nan < ρ + = r , for n > N, ⇒ an 6 r n.
n=
an =
n=
an +
n=N
an 6
n=
an +
n=N
r n
n=
an =
n=
an −
n=
r n^ +
n=
r n^ =
n=
(an − r n) +
1 − r
So
an is bounded. A non-decreasing, bounded above, series converges. The proofs for (b), (c) are similar to ratio test.
