Question

Find the interval of convergence

n!x^n

Answer 1

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:
\[R = \frac{1}{L} = \frac{1}{1} = 1\]

Answer 2

May I get a low level detail of the solution?

Answer 3

ok!

Answer 4

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

Answer 5

Thanks!

Answer 6

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

Answer 7

:)

Answer 8

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.

Answer 9

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!! :)

Answer 10

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

Answer 11

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 \(|x/r|<1\implies x\in (-r,r)\)

Answer 12

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 $$\lim\sqrt[n]{a_n} r\le 1\\\lim\sqrt[n]{a_n} \le \frac1r$$