# Thread: Some tough power series convergence vs divergence exercises

1. ## Some tough power series convergence vs divergence exercises

Any help/hints for any of these exercises would be much appreciated!

3. $\displaystyle \sum(-1)^n\binom{2n}{n}x^n$ has a radius of convergence of $\displaystyle \frac{1}{4}$. Determine whether or not each endpoint converges.

4. $\displaystyle \sum\frac{n^n}{n!}x^n$ has a radius of convergence of $\displaystyle \frac{1}{e}$. Determine whether or not each endpoint converges.

5. $\displaystyle \sum\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)}{\beta(\beta+1)\cdots(\beta+n-1)}x^n$ (with $\displaystyle \alpha,\beta\neq$ negative integer) has a radius of convergence of $\displaystyle 1$. Determine whether or not each endpoint converges.

Determining the radiuses of convergence for these series was easy, but I don't seem to have enough tools in my toolbox of math tricks to routinely solve the endpoints. I am therefore looking to learn new tricks! Please share them if you've got them.

Thanks!

EDIT: The following have been solved:

1. $\displaystyle \sum\left(1+\frac{1}{n}\right)^{n^2}[2+(-1)^n]^nx^n$ has a radius of convergence of $\displaystyle \frac{1}{3e}$. Determine whether or not each endpoint converges.

2. $\displaystyle \sum\left(2+\sin\frac{n\pi}{6}\right)^nx^n$ has a radius of convergence of $\displaystyle \frac{1}{3}$. Determine whether or not each endpoint converges.

2. 1. If the limit of the n'th term at the end point is not zero then the series diverges there. I thought this was an interesting limit to determine if it tends to zero:

$\displaystyle \lim_{n\to\infty} \frac{(1+\frac{1}{n})^{n^2}(2+(-1)^n)^n}{(3e)^n}=\lim_{n\to\infty}\frac{(1+1/n)^{n^2}}{e^n} \frac{(2+(-1)^n)^n}{3^n}$

3. 1. The coefficients of even terms of the series approaches $\displaystyle 3e^n$. Hence at both endpoints, the even terms of the series approach 1. Thus the terms fail to converge to zero, and so the series is divergent.

2. This is similar. Note that $\displaystyle 2 + \sin (\frac{n\pi}{6}) = 3$ for infinitely many n (namely, for n = 3, 15, 27, ...). Therefore at the endpoints $\displaystyle x = \frac{1}{3}, -\frac{1}{3}$, every twelfth term in the series is 1 or -1. So the terms don't go to zero, and the series is divergent.

I'll try to look over the others later.

4. (3) To determine if $\displaystyle a_n\to 0$, I used Stirlings Approximation $\displaystyle n!\sim n^n e^{-n}\sqrt{2\pi n}$

$\displaystyle \binom{2n}{n} \frac{1}{4^n}=\frac{(2n)!}{(n!)^2 4^n}\sim \frac{(2n)^{2n} e^{-2n}\sqrt{4 \pi n}}{(n^n e^{-n} \sqrt{2\pi n})^2 4^n}=\frac{\sqrt{4\pi n}}{2\pi n}$

and that tends to zero so the series may converge at the end points. At the right end point $\displaystyle x=1/4$, the series is alternating so I can use the Alternating Series Test and consider:

$\displaystyle \frac{a_{n+1}}{a_n}=\frac{2(n+1)(2(n+1)-1)(2n)!}{((n+1)n!)^2 4(4^n)}\frac{(n!)(n!) 4^n}{(2n)!}=\frac{4(n+1)^2-2(n+1)}{4(n+1)^2}$

which is less than one so the series converges at this point. At the left end-point we have

$\displaystyle \sum \frac{(2n)!}{(n!)^2}\frac{1}{4^n}$ and if I didn't make any mistakes, both the ratio test and root test give values of one so are not applicable.

However, if the terms of the series asymptotically approach $\displaystyle \frac{1}{\sqrt{\pi n}}$ then for n sufficiently large:

$\displaystyle \frac{(2n)!}{(n!)^2 4^n}>\frac{1}{(\pi n)^{3/4}}$ and we know $\displaystyle \sum \frac{1}{n^p}$diverges for $\displaystyle 0<p\leq 1$, thus the series diverges at the point $\displaystyle x=-1/4$.

5. Hey Hatoff, I think the convergence of (4) can be studied the same way I approached (3) above: Rely on Stirling's approximation, the Alternating Series Test, and the p-series comparison. When I do that, I get the region of convergence to be $\displaystyle \left[-\frac{1}{e},\frac{1}{e}\right)$. Also, Mathematica confirms the divergence at 1/e but cannot determine the convergence at -1/e.

However, this is the first time I've studied convergence at the boundaries so either someone need to point out any mistakes I've made or you'll need to check everything ok.

6. Well, those last three are definitely tricky. I am very interested in trying out Stirling's approximation in future exercises, but I am uncomfortable with using it in proofs like these.

Thanks for the help, everyone!