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The series:

$$\sum_{n=1}^\infty \left[\frac{1\cdot 3\cdot 5 \ldots (2n-1)}{2\cdot4\cdot6\ldots(2n)}\right]^p\cdot\left(\frac{x-1}{2}\right)^n$$

Therefore:

$$a_n=\left[\frac{1\cdot3\cdot5\ldots(2n-1)}{2\cdot4\cdot6\ldots(2n)}\right]^p\cdot\left(\frac{1}{2}\right)^n$$

I think it is possible to use ratio-like test:

$$\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right| \color{red} = \ldots \color{red} = 2\cdot\lim_{n \to \infty}\left(\frac{2n+1}{2n+2}\right)^p = 2 $$

Hence $\mathbf{R}=2$

Divergence interval:

From general defenition of power series we have $x_0=1$:

$$\sum_{n=1}^\infty=c_n(x-x_0)^n$$

And using interval defenition we have:

$$x \in (x_0-R;x_0+R) \Rightarrow x \in (-1;3)$$

But how shoul I check behaviour at the end points of the interval?

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  • $\begingroup$ Plug them in. For example x = 3 you end up with the series $\sum \frac{(2n)!}{2^{2n}(n!)^2}$. Use Stirling's formula to check the limit of the general term. $\endgroup$ – OR. Jun 16 '17 at 10:54
  • $\begingroup$ @MlazhinkaShungGronzalezLeWy so I just have to substitute both $x$ values into the function and check series convergence/divergence $\endgroup$ – M.Mass Jun 16 '17 at 12:41
  • $\begingroup$ Yes. It is just that often times the convergence at those points is more difficult to establish. Inside the radius of convergence comparison with a geometric series always works. At the boundary it never does. For example, it seems to be that in this problem one needs to compare with $1/\sqrt{n}$, if Matsuda's computations are correct, which is what he got when using Stirling's. For school problems, if I didn't know Stirling's or don't remember it I would try comparinson with one of the harmonic series $\sum n^{-\alpha}$ and see if some $\alpha$ is good to do the work. $\endgroup$ – OR. Jun 16 '17 at 14:28
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At $x=-1$, the series to be evaluated is $$ \sum_{n=1}^{\infty} (a_n)^p (-1)^n .$$ What the Stirling's formulas shows is the behavior of $n!$ when $n \to \infty$. More precicely it tells us that $$\lim_{n\to \infty} \frac{n!}{\sqrt{2\pi n}\, n^n e^{-n}} = 1, \quad \mbox{or} \quad n! \sim \sqrt{2 \pi n}\, n^n e^{-n}.$$ Applying this to $a_n$ we get $$ a_n = \frac{(2n)!}{2^{2n} (n!)^2} \sim \frac{\sqrt{2\pi (2n)}\, (2n)^{2n} e^{-2n}}{2^{2n} (\sqrt{2 \pi n}\, n^n e^{-n})^2} = \frac{1}{\sqrt{\pi n}}.$$ Thus $a_n \to 0$ when $n \to \infty$. Therefore the alternating series $\sum (a_n)^p (-1)^n$ converges if $p>0$. Of course this series does not converge when $p \leq 0$. At $x=3$, the sum $\sum (a_n)^p$ converges if and only if $p>2$. This is because the sum $\sum (1/n^q)$ converges if and only if $q>1$.

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