1. ## Interval of convergence

I need some help on the following

Determine the interval of convergence of the power series

infinity
sum of series (-1)^n/(n+3)4^n * (x-1)^n
n=1

I have

Here An = (-1)^n/(n+3)4^n, for n=1,2....

1. since

|An+1/An| = 1/(n+4)4^n+1 * (n+3)4^n/1

= ? ----> as n ----> ?

we have R = ? , by the Ratio Test.

Thus

2. If x = ? , the the power series is

?

Hence the interval of convergence is ( ? )

2. ## Re: Interval of convergence

Is this your mid-term quiz? You should be demonstrating SOME knowledge fo the subject. You seem to be just guessing.

3. ## Re: Interval of convergence

your series is not real since $\displaystyle (-1)^\frac{n}{n+3}$ is imaginary for n as odd

4. ## Re: Interval of convergence

Originally Posted by waqarhaider
your series is not real since $\displaystyle (-1)^\frac{n}{n+3}$ is imaginary for n as odd
You have miss read the question.
It is $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{(n + 3)4^n }}(x - 1)^n } ~.$

Consider the root test.
$\displaystyle \frac{{\left| {x - 1} \right|}}{{\sqrt[n]{{n + 3}}(4)}} \to ?$

5. ## Re: Interval of convergence

I am not sure about the root test, mainly been working on the Ratio test.

6. ## Re: Interval of convergence

Originally Posted by Arron
I am not sure about the root test, mainly been working on the Ratio test.
Essentially the two test are in most cases equivalent.
They both rely upon the same conclusion
If $\displaystyle \lim _{n \to \infty } {{\sqrt[n]{{\left|a_n\right| }}} } < 1$ the series converges.

7. ## Re: Interval of convergence

So does that tend to 1 as n ----> infinity

8. ## Re: Interval of convergence

Originally Posted by Arron
So does that tend to 1 as n ----> infinity
NO!
$\displaystyle \frac{{\left| {x - 1} \right|}}{{\sqrt[n]{{n + 3}}(4)}} \to \frac{{\left| {x - 1} \right|}}{4}$

Now solve $\displaystyle {\frac{|x-1|}{4}<1$.

Be sure to test the endpoints.

9. ## Re: Interval of convergence

Sorry I find this really hard. Would it be 1/4. Thanks for all your help.

10. ## Re: Interval of convergence

Originally Posted by Arron
Sorry I find this really hard. Would it be 1/4. Thanks for all your help.
Would what be 1/4?? Please quote the replies you are refering to when asking further help.

11. ## Re: Interval of convergence

By using d'Alembert test you will come to the same conclusion. We have to calculate (in general):
$\displaystyle \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=r$
So in this case:
$\displaystyle \lim_{n\to \infty} \left|\frac{\frac{(-1)^{n+1}\cdot (x-1)^{n+1}}{(n+4)\cdot 4^{n+1}}}{\frac{(-1)^n\cdot (x-1)^n}{(n+3) /cdot 4^n}}\right |$
$\displaystyle =\lim_{n \to \infty} \left | \frac{(-1)^{n+1}\cdot (x-1)^{n+1}\cdot (n+3)\cdot 4^n}{(n+4)\cdot 4^{n+1}\cdot (-1)^n\cdot (x-1)^n}\right|$
$\displaystyle =\lim_{n \to \infty} \left |(-1)\cdot (x-1)\cdot \frac{1}{4}\cdot \frac{n+3}{n+4}\right |$
$\displaystyle =\frac{|x-1|}{4}\lim_{n \to \infty} \left|\frac{n+3}{n+4}\right |=\frac{|x-1|}{4}$

Along the test the serie wil converge if $\displaystyle 0<\frac{|x-1|}{4}<1$ and diverge if $\displaystyle \frac{|x-1|}{4}>1$ so you've to calculate the $\displaystyle x$ values wherefore:
$\displaystyle \frac{|x-1|}{4}<1\Leftrightarrow |x-1|<4$
So two different cases:
(1) $\displaystyle x-1<4 \Leftrightarrow ...$
(2) $\displaystyle -x+1<4 \Leftrightarrow ...$