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

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here 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: \[R = \frac{1}{L} = \frac{1}{1} = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0May I get a low level detail of the solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2your power series doesn't converge at x=1 or x=1 so it converges into the interval: (1, 1)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2of course at x=0, your series converges, as you can easily check, by substitution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Was 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.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I'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!! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the series has zero radius of convergence because eventually \(n!\) grows much faster than \(x^{n}\) can shrink

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0essentially, 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 \(x/r<1\implies x\in (r,r)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we 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 $$\lim\left\frac{a_n}{r^{n}}\right\le1\\\lim \lefta_n r^n\right=\lim(\sqrt[n]{a_n} r)^n=(\lim\sqrt[n]{a_n} r)^n\le1$$ so $$\lim\sqrt[n]{a_n} r\le 1\\\lim\sqrt[n]{a_n} \le \frac1r$$
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