# radius of convergence of a power series

These are questions given for an assignment and I REALLY need to get a good grade so please could you guys check over them and give some pointers where I've gone wrong?

thanks.

ed: I hope it's ok to overwrite the original post:

$$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}}$$

\begin{align} a_n &= \frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}\\ a_{n+1} &= \frac{(2x-8)^{n+1}}{4^{n+1}((n+1)^{2}+2)}.\\ \left|\frac{a_n}{a_{n+1}}\right| &= \left|\frac{(2x-8)^{n+1}}{4^{n+1}((n+1)^{2}+2)} \times \frac{4^{n}(n^{2}+2)}{(2x-8)^{n}}\right|.\\ \lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|&= |2x-8| \lim_{n \to \infty}\frac{n^{2}+2}{4(n^{2}+2n+3)}\\ &= |2x-8| \times\frac{1}{4}. \end{align} Therefore $L = \frac{1}{4}$ and series converges when $|2x-8| \times\frac{1}{4} < 1$.

Radius of convergence is $2$.

Endpoints; $|2x-8| = 4 \rightarrow x = 6$ and $x=2$.

Testing endpoints. When $x = 6$: $$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}} = \sum_0^\infty \frac{1}{n^{2}+2}.$$ convergent by comparison test with $\sum\frac{1}{n^{2}}$.

When x = 2: $$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}} = \sum_0^\infty\frac{(-1)^n}{n^{2}+2},$$ convergent by alternating sequence test and is absolutely convergent.

Therefore the series $$\sum_0^\infty {\frac{(2x-8)^{n}}{4^{n}(n^{2}+2)}}$$ is convergent for $x\in [2,6]$.

2)b

$$\sum_1^\infty\frac{(-1)^n}{n4^{2n}}(x+3)^n$$

\begin{align} a_n &= \frac{(-1)^n}{n4^{2n}},\\ a_{n+1} &= \frac{(-1)^{n+1}}{(n+1)4^{2(n+1)}}.\\ \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{(-1)^{n+1}}{(n+1)4^{2(n+1)}} \times \frac{n4^{2n}}{(-1)^n}\right|\\ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &= |x+3|\lim_{n \to \infty}\frac{n}{4^2(n+1)}\\ &= |x+3|\times\frac{1}{16}. \end{align} Series converges when $|x+3|\times\frac{1}{16} \lt 1$. Radius of convergence \is $16$.

Endpoints: $|x+3| = 16 \rightarrow x = 13$ or $-19$.

Testing endpoints: When $x = 13$, $$\sum_1^\infty\frac{(-1)^n(16)^n}{n4^{2n}} = \sum_1^\infty\frac{(-1)^n)}{n}.$$ This is an alternating harmonic series and therefore convergent.

When x = -19, $$\sum_1^\infty\frac{(-1)^n(-16)^n}{n4^{2n}}{} = \sum_1^\infty\frac{(-1)^{2n}}{n}.$$ Harmonic series and therefore divergent.

Therefore the series $$\sum_1^\infty\frac{(-1)^n}{n4^{2n}}(x+3)^n$$ is convergent for $x \in(-19, 13]$.

2) c)

$$\sum_0^\infty\frac{(1+3^{n-1})x^n}{(n+1)!}$$ \begin{align} a_n &= \frac{(1+3^{n-1})x^n}{(n+1)!}\\ a_n+1 &= \frac{(1+3^{(n+1)-1})x^{n+1}}{((n+1)+1)!}.\\ \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{(1+3^n)x^{n+1}}{(n+2)!} \times \frac{(n+1)!}{(1+3^{n-1})x^n}\right|\\ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &= |x|\lim_{n \to \infty} \frac{1+3^n}{1+3^{n-1}} \times \lim_{n \to \infty} \frac{1}{n+2}\\ &= |x|\lim_{n \to \infty} \frac{3^{-n}+ 1}{3^{-n}+3^{-1}} \times 0\\ &= |x|\lim_{n \to \infty} \frac{0 + 1}{0 +3^{-1}} \times 0\\ &= |x|\times3\times0 = 0. \end{align}

Therefore, $L = 0$ and the series converges absolutely for all $x$. The radius of convergence is $\infty$.

$\displaystyle\sum_0^\infty\frac{(1+3^{n-1})x^n}{(n+1)!}$ converges absolutely for $x\in(-\infty,\infty)$.

Cheers,

Gregg

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In working the first problem, you slipped when stating the radius of convergence is 4. The radius of convergence is the distance of x from the point around which the power series expansion is taken. Note $2x-8 = 2(x-4)$. You found the right endpoints, $x=2,6$. How far does $x$ get from the center of the expansion, $x=4$ ? –  hardmath Dec 21 '10 at 13:02
Hi hardmath, thanks for pointing the error with the radius of convergence. –  gnicezw Dec 22 '10 at 5:40
Not every math equation has to be in math display; you can also use in-line formulas, by using $...$ instead of $$...$$. For long series of equations, try to align them at the equal sign, using \begin{align} and \end{align} instead of a series of displays (use & to mark the point of alignment, \\ to mark the end of lines (should be just \ but currently this confuses the renderer). The sign for membership is \in, not \epsilon. And if you must put text in a math equation, use \text{...}` for it. –  Arturo Magidin Dec 22 '10 at 21:47

It seems mostly OK (at a quick glance), but there are few minor issues.

Don't write $|a_n/a_{n+1}|=(\dots)=\lim (\dots)$, because the elements of the sequence are not equal to the limit of the sequence. Either you write "lim" in each step, like $\lim (\dots) =\lim (\dots) = A$, or else use the arrow notation $(\dots) = (\dots) \to A$. (But don't mix the two notations as many people do; that is, don't write "$\lim(\dots) \to A$"!).

Write $x\in [2,6]$, not $x=(2,6)$. Also: $x \in (-19,13]$ (the smallest number first).

In the last problem, you've made a mistake: $(n+1)!/(n+2)! = 1/(n+2)$, not $(n+1)/(n+2)$.

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Hi Hans, Thanks for the respons. I had been wondering about the validity of ratio = (. . . ) Lim(. . . ). Thanks for the suggestions it's late here so will have a few hours sleep and then work on your suggestions :) –  gnicezw Dec 21 '10 at 13:38

In the first problem, you have the wrong radius of convergence. You forgot that you have a factor of $2$ in the term $|2x-8|$. Second problem is done correctly.

For the last problem, I would suggest doing the arithmetic correctly. You have a mistake in the computation of the limits (as well as, though it is moot, a mistake in the computation of the series when $x=\frac{1}{3}$).

I was going to mention the issues Hans Lundmark raised, about confusing the terms of the sequence with the limits and the value of the variable with the interval in which it may lie for convergence.

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Hi Arturo, Thank you also for your input. will update post in a few hours. cheers, Gregg –  gnicezw Dec 21 '10 at 13:40
Hi, Thanks for all your help :) . I have updated the reworked solutions and agreed; my understanding of factorials and series, sequences and their limits is poor. I've had to prioritise the assignment :( and will re-read proofs over the xmas break. cheers, Gregg –  gnicezw Dec 22 '10 at 5:49
@gnicezw: and when you updated, you also undid all of my careful formatting that made it more intelligible. You might want to try imitating it next time. –  Arturo Magidin Dec 22 '10 at 21:35