Find the radius of convergence of $\sum_{n=1}^\infty n!(2x-1)^n$
Now, by D'Alemberts Ratio test that implies (for convergence):
$\lim_{n\to\infty} \lvert \frac{(n+1)!(2x-1)(2x-1)^n}{n!(2x-1)^n} \rvert<1$
$\lim_{n\to\infty} \lvert (n+1)(2x-1)\rvert<1 $
$\lvert 2x-1\rvert<\lim_{n\to\infty} \lvert\frac{1}{(n+1)}\rvert$
$\lvert 2x-1\rvert < 0$
$\lvert 2x\rvert< 1$
$\frac{-1}{2}<x<\frac{1}{2}$
Therefore radius of convergence $(R) = \frac{1}{2}$
However upon inspection this answer is incorrect since for a value of $x = 0$, the sequence clearly diverges.
Can someone tell me where I have gone wrong?
Thanks a million!
Best Answer
Assuming you mean
$$\sum_{n=1}^\infty n!(2x-1)^n=\sum_{n=1}^\infty n!2^n(x-0.5)^n$$
Now use ratio test:
$$\frac{|a_{n+1}|}{|a_n|}=\frac{(n+1)!2^{n+1}}{n!2^n}=2(n+1)>1$$ Inequality holds for every $n$. Hence the radius of convergence is $0$.
Formally you have to calculate
$$r=\lim_{n\to \infty}\frac{1}{\frac{|a_{n+1}|}{|a_n|}}=\lim_{n\to \infty}\frac{1}{2(n+1)}=0$$
Edit: Your method is wrong, because: $$\lvert 2x-1\rvert < 0$$ This inequality is can never be true for any $x$ as the absolute value is never negative.