## anonymous one year ago Find the interval of convergence n!x^n

1. Michele_Laino

here we have a power series, so we have to apply this formula: $R = \frac{1}{L}$ where R is the converging radius and L is: $L = \mathop {\max \lim }\limits_{n \to \infty } \sqrt[n]{{\left| {{a_n}} \right|}} = \mathop {\max \lim }\limits_{n \to \infty } \sqrt[n]{{n!}} = 1$ so, we have: $R = \frac{1}{L} = \frac{1}{1} = 1$

2. anonymous

May I get a low level detail of the solution?

3. Michele_Laino

ok!

4. Michele_Laino

your power series doesn't converge at x=1 or x=-1 so it converges into the interval: (-1, 1)

5. anonymous

Thanks!

6. Michele_Laino

of course at x=0, your series converges, as you can easily check, by substitution

7. Michele_Laino

:)

8. anonymous

Was it $\sum_{n=0}^{\infty} n! x^n$ if so the radius of convergence then would be R = 0, where the interval of convergence is {0}. We could use the ratio test, we must note that $x \neq 0$ so let $a_n = n!x^n$$\lim_{n \rightarrow \infty} \left| \frac{ a_n+1 }{ a_n } \right| = \lim_{n \rightarrow \infty} \left| \frac{ (n+1)!x^{n+1} }{ n!x^n } \right| = \lim_{n \rightarrow \infty} (n+1)|x| = \infty$ xo you can even see the series diverges when x cannot = -, so the series converges only when x = 0.

9. Michele_Laino

I'm very sorry @Yaros, I have made an error, here is the right computation: $\Large \begin{gathered} L = \mathop {\max \lim }\limits_{n \to \infty } \sqrt[n]{{\left| {{a_n}} \right|}} = \mathop {\max \lim }\limits_{n \to \infty } \sqrt[n]{{n!}} = + \infty \hfill \\ \hfill \\ R = 0 \hfill \\ \end{gathered}$ Please look at the correct answer of @iambatman Thanks @iambatman for your reply!! :)

10. anonymous

the series has zero radius of convergence because eventually $$n!$$ grows much faster than $$x^{-n}$$ can shrink

11. anonymous

essentially, convergence of a series $$\sum a_n x^n$$ depends on $$|a_n|<r^{-n}$$ such that $$\sum a_n x^n<\sum (x/r)^n$$ which converges when $$|x/r|<1\implies x\in (-r,r)$$

12. anonymous

we try to find the maximum such $$r$$, so we want to match the growth of $$a_n$$: $$a_n\in\Theta(r^{-n})$$ if we know that $$a_n\sim r^{-n}$$ then it follows we want $$\lim\left|\frac{a_n}{r^{-n}}\right|\le1\\\lim \left|a_n r^n\right|=\lim(\sqrt[n]{a_n} r)^n=(\lim\sqrt[n]{a_n} r)^n\le1$$ so $$\lim\sqrt[n]{a_n} r\le 1\\\lim\sqrt[n]{a_n} \le \frac1r$$