MATH1051 Lecture Notes - Lecture 21: Nth Root, Ratio Test
Lecture #21 – Ratio and Root Tests
Geometric Sequences
●r r r r r .. r r∑
∞
n=1
a1n−1 =a1+a1+a12+a13+a14+. +a1n−1 +a1n
●Ratio Test
○ r
an
an+1 =
○We can find the ratio of the geometric series this way
●lim
n→∞ an
an+1 =r
●Root Test
○ r
√
nan=√
na r
1n−1 = r
√
na1
●As long as we take the , we can take the nth root to the nth termlim
n→∞
●Take two different ways to find the essence of geometric series
○Let or R= lim
n→∞
|
|an
an+1 |
| R= lim
n→∞ √
na
|n|
○Then if the geometric series is convergent 1,R> ∑
an
○Or if , the series is divergent 1R>
○? 1R=
■This means that there’s no information to be had about because∑
an
there is no geometric growth or decay
■In this case, we’d resort to using a different test
Example #1
●∑
∞
n=1
n
2n
○Let’s try both tests
○Ratio Test
■R lim
n→∞
|
|an
an+1 |
| = = lim
n→∞
|
|
|n
2n
n+1
2n+1 |
|
|
■( )lim
n→∞
n+1
2n+1 n
2n
●Divide
■lim
n→∞ 2 (n)
n+!
2 (n+1)
n
●You can cancel out several terms here after placing the terms
parallel for easier cancelling
■ lim
n→∞ 2n
n+1 = 2
1
■, therefore, convergent 1R<
○Root Test
■ lim
n→∞ √
na
|n|
■ lim
n→∞ √
nn
2n
■lim
n→∞
√
nn
√
n2n
●Here, we separate the two terms on the top and bottom
■lim
n→∞ 2
1
●Because , and , it makes the limit that much easier
√
nn= 1 √
n2n= 2
■2
1
lim
n→∞ √
np(n) = 1
●For every polynomial p, that limit is equal to 1
Example #2
●∑
∞
n=1 n
1
○We already know that this is divergent because of p-test, but we can try ratio
and root tests to see how it works out
○Ratio: lim
n→∞
|
|
|n
1
1
n+1 |
|
|
Document Summary
Geometric sequences n=1 a1 r n 1 = a1 + a1 + a1 r r. + a1 r n 1 + a1 n r. We can find the ratio of the geometric series this way an+1 = r. As long as we take the lim n . , we can take the nth root to the nth term n 1 = r. Take two different ways to find the essence of geometric series. This means that there"s no information to be had about an because there is no geometric growth or decay. In this case, we"d resort to using a different test. You can cancel out several terms here after placing the terms parallel for easier cancelling. Root test n n a| n| lim n n n lim. Here, we separate the two terms on the top and bottom. , it makes the limit that much easier. 2 n n p(n) = 1 lim.
