MHB Series Convergence and Divergence III

[ardentmed]

Hey guys,

I have a few quick questions for the problem set I'm working on at the moment:
View attachment 2774
I'm highly doubtful of my answer for c. I used the roots test instead of the ratio test, which gives 1/n, which I took the limit of to get an interval of [-∞ , ∞]

As for a and b, I got [-5,5] and (-∞, ∞) respectfully.

Thanks in advance.

 

[evinda]

Hey guys,

I have a few quick questions for the problem set I'm working on at the moment:
https://www.physicsforums.com/attachments/2774
I'm highly doubtful of my answer for c. I used the roots test instead of the ratio test, which gives 1/n, which I took the limit of to get an interval of [-∞ , ∞]

As for a and b, I got [-5,5] and (-∞, ∞) respectfully.

Thanks in advance.

The general formula of a Taylor Series is $\sum_{n=0}^{\infty} a_n(x- \xi)^n$

At part $c$,it is: $a_n=\frac{(-1)^n}{n2^n} , \xi=-2$

$$ \rho=\lim_{n \to +\infty} \sqrt[n]{|a_n|}= \lim_{n \to +\infty} \sqrt[n]{\frac{1}{n2^n}}=\frac{1}{2}$$

$$R=\frac{1}{ \rho}=2$$

So,the series converges absolutely for $x \in (\xi-R, \xi+R)=(-4,0)$ and diverges for $x \notin [-4,0]$

So,now it remains to check if the series converges for $x=-4 \text{ and for } x=0$.

 

[Prove It]

Hey guys,

I have a few quick questions for the problem set I'm working on at the moment:
https://www.physicsforums.com/attachments/2774
I'm highly doubtful of my answer for c. I used the roots test instead of the ratio test, which gives 1/n, which I took the limit of to get an interval of [-∞ , ∞]

As for a and b, I got [-5,5] and (-∞, ∞) respectfully.

Thanks in advance.

Why not do the ratio test for (c) as well?

$\displaystyle \begin{align*} \lim_{n \to \infty} \frac{ \left| a_{n + 1} \right| }{\left| a_n\right| } &< 1 \\ \lim_{n \to \infty} \frac{\left| \left( -1 \right) ^{n+1} \frac{\left( x + 2 \right) ^{n+1}}{\left( n + 1 \right) \, 2^{n+1}} \right| }{\left| \left( -1 \right) ^n \frac{\left( x + 2 \right) ^n}{n\,2^n} \right| } &< 1 \\ \lim_{n \to \infty} \frac{\frac{\left| x + 2 \right| ^{n + 1} }{\left( n + 1 \right) \, 2^{n + 1} }}{\frac{\left| x + 2 \right| ^n }{n\,2^n}} &< 1 \\ \lim_{n \to \infty} \frac{n\,2^n \, \left| x + 2 \right| ^{n + 1}}{\left( n + 1 \right) \, 2^{n + 1} \, \left| x + 2 \right| ^n } &< 1 \\ \frac{ \left| x + 2 \right| }{2} \, \lim_{n \to \infty} \frac{n}{n + 1} &< 1 \\ \frac{ \left| x + 2 \right| }{2} \, \lim_{n \to \infty} \left( 1 - \frac{1}{n + 1} \right) &< 1 \\ \frac{\left| x + 2 \right| }{2} \cdot 1 &< 1 \\ \frac{ \left| x + 2 \right| }{2} &< 1 \\ \left| x + 2 \right| &< 2 \\ -2 < x + 2 &< 2 \\ -4 < x &< 0 \end{align*}$

So your radius of convergence is 2, and your integral of convergence may be -4 < x < 0, but you will also need to check the endpoints.