Determine the values of x for series convergence

[adomad123]

[itex][/itex]

Homework Statement

Determine the values of x for which the following series converges. Remember to test the end points of the interval of convergence.
[itex]^{∞}_{n=0}\sum\frac{(1-)^{n+1}(x+4)^{n}}{n}[/itex]

Homework Equations

I worked it down to
|x+4|<1
∴-5<x<-3

The Attempt at a Solution

When I come to test the end point when x=-3

Ʃ=[itex]\frac{(-1)^{n+1}(-1)^{n}}{n}[/itex]
ALTERNATING SERIES
Test 1. See if [itex]lim_{n→∞}a_{n}[/itex]=0

[itex]lim_{n→∞}[/itex][itex]\frac{(1)^{n}}{n}[/itex]
=[itex]lim_{n→∞}[/itex][itex]\frac{(1)}{n}[/itex] =0

test 2. See if [itex]a_{n+1}<a_{n}[/itex]

[itex]a_{n+1}= \frac{1}{n+1}<\frac{1}{n}[/itex]
∴[itex]a_{n+1}<a_{n}[/itex]

hence, series converges when x=-3


When x=-5
Ʃ=[itex]\frac{(-1)^{n+1}(-1)^{n}}{n}[/itex]
Test 1 see if [itex]lim_{n→∞}a_{n}[/itex]=0\

[itex]lim_{n→∞}[/itex][itex]\frac{(-1)^{n}}{n}[/itex]
=[itex]lim_{n→∞}[/itex][itex]\frac{\frac{(-1)^{n}}{n}}{1}[/itex] =0


Test 2. [itex]a_{n+1}<a_{n}[/itex]
[itex]a_{n+1}[/itex]=[itex]\frac{-(-1)^{n}}{n+1}<\frac{(-1)^{n}}{n+1}[/itex] (since negative of the other?

Hence, series converges when x=-5

∴ series converges when -5≤x≤-3

Not sure if I have the right answer. but I don't know what to do when I get [itex]a_{n}[/itex] has the [itex](-1)^{n}[/itex]

 

[HallsofIvy]

[itex][/itex]

Homework Statement

Determine the values of x for which the following series converges. Remember to test the end points of the interval of convergence.
[itex]^{∞}_{n=0}\sum\frac{(1-)^{n+1}(x+4)^{n}}{n}[/itex]

Homework Equations

I worked it down to
|x+4|<1
∴-5<x<-3

The Attempt at a Solution

When I come to test the end point when x=-3

Ʃ=[itex]\frac{(-1)^{n+1}(-1)^{n}}{n}[/itex]

(-3+ 4)= 1, not -1. You should not have the [itex](-1)^n[/itex] term.
However, that means the series is an alternating series so what you have is correct. See below for the x= -5 case.

ALTERNATING SERIES
Test 1. See if [itex]lim_{n→∞}a_{n}[/itex]=0

[itex]lim_{n→∞}[/itex][itex]\frac{(1)^{n}}{n}[/itex]
=[itex]lim_{n→∞}[/itex][itex]\frac{(1)}{n}[/itex] =0

test 2. See if [itex]a_{n+1}<a_{n}[/itex]

[itex]a_{n+1}= \frac{1}{n+1}<\frac{1}{n}[/itex]
∴[itex]a_{n+1}<a_{n}[/itex]

hence, series converges when x=-3


When x=-5
Ʃ=[itex]\frac{(-1)^{n+1}(-1)^{n}}{n}[/itex]

[itex](-1)^{n+1}(-1)^n= (-1)^{2n+1}= (-1)^{2n}(-1)= -1[/itex]
The sum is just [itex]-\sum\frac{1}{n}[/itex] which does NOT converge

Test 1 see if [itex]lim_{n→∞}a_{n}[/itex]=0\

[itex]lim_{n→∞}[/itex][itex]\frac{(-1)^{n}}{n}[/itex]
=[itex]lim_{n→∞}[/itex][itex]\frac{\frac{(-1)^{n}}{n}}{1}[/itex] =0


Test 2. [itex]a_{n+1}<a_{n}[/itex]
[itex]a_{n+1}[/itex]=[itex]\frac{-(-1)^{n}}{n+1}<\frac{(-1)^{n}}{n+1}[/itex] (since negative of the other?

Hence, series converges when x=-5

∴ series converges when -5≤x≤-3

Not sure if I have the right answer. but I don't know what to do when I get [itex]a_{n}[/itex] has the [itex](-1)^{n}[/itex]

 

[adomad123]

mate, you're a legend!