# Determine the values of x for series convergence

1. Nov 5, 2013

1. The problem statement, all variables and given/known data

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

2. Relevant equations

I worked it down to
|x+4|<1
∴-5<x<-3

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 $a_{n+1}<a_{n}$

$a_{n+1}= \frac{1}{n+1}<\frac{1}{n}$
∴$a_{n+1}<a_{n}$

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. $a_{n+1}<a_{n}$
$a_{n+1}$=$\frac{-(-1)^{n}}{n+1}<\frac{(-1)^{n}}{n+1}$ (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 $(-1)^{n}$

Last edited: Nov 5, 2013
2. Nov 5, 2013

### HallsofIvy

Staff Emeritus
(-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.

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

3. Nov 5, 2013