Summary
Summary Ratio Test | Calculus II Notes
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- Institution
Highlights theorems and gives detailed and explained examples on the ratio test
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3.3.5 The Ratio TestThe ratio test series
&
is based on a reexamination ofthe
geometric
·
An nE, ar converges
=
when Ir/1 and diverges otherwises
the series is based
convergence ofthe on a
completely.
↓ is justthe ratio of successive terms:
-
r An+1
=
an
Theorem:
letN be
any positive integer, and assume an F0 for all n=N
if i Ants L1, E, An converges
=
an
E, An diverge
him ant-
if or him Anti- +,
n 0
an
=
The ratio test provides no conclusion if
an
example: anx-
a, x real numbers
->
any nonzero
we
&
have seen thatthe
geometric series ax" converges
,
when lxx1 and diverges when xc
Ean, anxn-1
An=
will never be to so you can pull itoutfrom the absolute value
An+1 a(n+DX
= = n +1(x)
n
(1
=
+
x) 2
=
1X) as
= n=
an xx-1
an
the ratio testtells us that Eranx" "converges if Ix/1 and diverges iflxK) says nothing
but in the
cases ofX =
11, but
by divergence test in both cases an=an(11)" does not
converg to zero
as nto.... diverges.
it
example: no ni, Xn+
san, an
n,
=
Xn+
at x x
= 1
=
x
+
=1Xas nee
1
+
an
converges if IX)<1 and diverges if X<1
if X 11? =
the series reduces to
o Xn+ X 1
=
-ona where m n+ =
this is simply a time the harmonic series which diverges
o n+ , Xn+ x =
- 1
=Eo(-1n+
which converges by the alternating series test
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