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anonymous

  • one year ago

Find the interval of convergence n!x^n

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  1. Michele_Laino
    • one year ago
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    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
    • one year ago
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    May I get a low level detail of the solution?

  3. Michele_Laino
    • one year ago
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    ok!

  4. Michele_Laino
    • one year ago
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    your power series doesn't converge at x=1 or x=-1 so it converges into the interval: (-1, 1)

  5. anonymous
    • one year ago
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    Thanks!

  6. Michele_Laino
    • one year ago
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    of course at x=0, your series converges, as you can easily check, by substitution

  7. Michele_Laino
    • one year ago
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    :)

  8. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    the series has zero radius of convergence because eventually \(n!\) grows much faster than \(x^{-n}\) can shrink

  11. anonymous
    • one year ago
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    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
    • one year ago
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    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 $$\lim\sqrt[n]{a_n} r\le 1\\\lim\sqrt[n]{a_n} \le \frac1r$$

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