Determine the values of x for series convergence

[adomad123]

Homework Statement

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

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

lim_{n→∞}\frac{(1)^{n}}{n}
=lim_{n→∞}\frac{(1)}{n} =0

test 2. See if a_{n+1}&lt;a_{n}

a_{n+1}= \frac{1}{n+1}&lt;\frac{1}{n}
∴a_{n+1}&lt;a_{n}

hence, series converges when x=-3


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

lim_{n→∞}\frac{(-1)^{n}}{n}
=lim_{n→∞}\frac{\frac{(-1)^{n}}{n}}{1} =0


Test 2. a_{n+1}&lt;a_{n}
a_{n+1}=\frac{-(-1)^{n}}{n+1}&lt;\frac{(-1)^{n}}{n+1} (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 a_{n} has the (-1)^{n}

 

[HallsofIvy]

Homework Statement

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

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=\frac{(-1)^{n+1}(-1)^{n}}{n}

(-3+ 4)= 1, not -1. You should not have the (-1)^n 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 lim_{n→∞}a_{n}=0

lim_{n→∞}\frac{(1)^{n}}{n}
=lim_{n→∞}\frac{(1)}{n} =0

test 2. See if a_{n+1}&lt;a_{n}

a_{n+1}= \frac{1}{n+1}&lt;\frac{1}{n}
∴a_{n+1}&lt;a_{n}

hence, series converges when x=-3


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

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

Test 1 see if lim_{n→∞}a_{n}=0\

lim_{n→∞}\frac{(-1)^{n}}{n}
=lim_{n→∞}\frac{\frac{(-1)^{n}}{n}}{1} =0


Test 2. a_{n+1}&lt;a_{n}
a_{n+1}=\frac{-(-1)^{n}}{n+1}&lt;\frac{(-1)^{n}}{n+1} (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 a_{n} has the (-1)^{n}

 

[adomad123]

mate, you're a legend!