Σ [(3n+2)/(3n-1)]^n where n=1, and infinity is on top of Σ
i did the root test and i got 1. so according to my teacher, 1=fails and i'm suppose to try another test.
(3n + 2)/(3n - 1) = (3n - 1 + 3)/(3n - 1) = 1 + 3/(3n - 1) > 1 for all n >= 1
Each term is a number greater than 1 to some power so is greater than 1.
All of the terms greater than 1 means that the infinite sum of them must itself be infinite so the series does not converge.
Have you tried the test of divergence?
If a series converges, then
lim (a[n]) = 0
n -> infinity
So the contrapositive of this statement is:
if lim (a[n]) is NOT 0, then the series diverges.
So let's try taking the limit of the general term as n approaches infinity
lim [ (3n + 2)/(3n - 1) ]^n
This is in the form [1^infinity], which is indeterminate.
Use the property [f(x)]^[g(x)] = e^[ g(x) ln( f(x) ) ]
lim [ e^[ n ln [ (3n + 2)/(3n - 1) ] ] ]
And now, we can move the limit INTO the exponent.
e^ [ lim ( n ln [ (3n + 2)/(3n - 1) ] ) ]
. . n -> infinity
And move the n to the denominator as (1/n).
e^ [ lim ( ln [ (3n + 2)/(3n - 1) ] / (1/n) ) ]
Now this is of the form [0/0]. Use L'Hospital's rule.
To skip details, the derivative of the log on the numerator is:
(-9)(3n + 2)/(3n - 1)
And the derivative of (1/n) is
(-1/n^2)
e^ [ lim ( (-9)(3n + 2)/(3n - 1) / (-1/n^2) ) ]
Cancellation of minus,
e^ [ lim ( (9)(3n + 2)/(3n - 1) / (1/n^2) ) ]
Reduction of complex fraction by multiplying top and bottom by n^2(3n - 1),
e^ [ lim ( (9)(3n + 2)(n^2) / (3n - 1) ]
The power of the numerator is bigger than the denominator, so we can determine using Calculus I concept that the limit is actually, so we get
e^0
which is
1
Therefore, the series diverges by the Test for Divergence.
Comments
(3n + 2)/(3n - 1) = (3n - 1 + 3)/(3n - 1) = 1 + 3/(3n - 1) > 1 for all n >= 1
Each term is a number greater than 1 to some power so is greater than 1.
All of the terms greater than 1 means that the infinite sum of them must itself be infinite so the series does not converge.
Have you tried the test of divergence?
If a series converges, then
lim (a[n]) = 0
n -> infinity
So the contrapositive of this statement is:
if lim (a[n]) is NOT 0, then the series diverges.
So let's try taking the limit of the general term as n approaches infinity
lim [ (3n + 2)/(3n - 1) ]^n
n -> infinity
This is in the form [1^infinity], which is indeterminate.
Use the property [f(x)]^[g(x)] = e^[ g(x) ln( f(x) ) ]
lim [ e^[ n ln [ (3n + 2)/(3n - 1) ] ] ]
n -> infinity
And now, we can move the limit INTO the exponent.
e^ [ lim ( n ln [ (3n + 2)/(3n - 1) ] ) ]
. . n -> infinity
And move the n to the denominator as (1/n).
e^ [ lim ( ln [ (3n + 2)/(3n - 1) ] / (1/n) ) ]
. . n -> infinity
Now this is of the form [0/0]. Use L'Hospital's rule.
To skip details, the derivative of the log on the numerator is:
(-9)(3n + 2)/(3n - 1)
And the derivative of (1/n) is
(-1/n^2)
e^ [ lim ( (-9)(3n + 2)/(3n - 1) / (-1/n^2) ) ]
. . n -> infinity
Cancellation of minus,
e^ [ lim ( (9)(3n + 2)/(3n - 1) / (1/n^2) ) ]
. . n -> infinity
Reduction of complex fraction by multiplying top and bottom by n^2(3n - 1),
e^ [ lim ( (9)(3n + 2)(n^2) / (3n - 1) ]
. . n -> infinity
The power of the numerator is bigger than the denominator, so we can determine using Calculus I concept that the limit is actually, so we get
e^0
which is
1
Therefore, the series diverges by the Test for Divergence.