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Consider the following series: $$ \sum _{n=0}^{\infty }\:n!\left(2x-1\right)^n $$ If I try to check for what $x$'s my series will converge, applying the ratio test, we'll see that $$ \lim _{n\to \infty \:}\left(\left|n+1\right|\right) = \infty\\ \left|2x-1\right|\cdot \infty \: = \infty $$ And because of that my serie is going to diverge for all $x$.

That's what I've concluded. But the answer is that the radius of convergence is going to be 0 and it'll converge for $x = \frac{1}{2}$.

How am I supposed to know that if a series diverges for all $x$ in the ratio test, there is still a possibility for it to converge for a $x$?! It's not making sense to me.

Thanks!

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    Hint: the root test tells you the radius of convergence when looking at a power series. – 2017-02-07
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    Sorry mate, I meant ratio test! sorry about that! I've corrected it – 2017-02-07
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    (above hint is still valid, please look at it) – 2017-02-07
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    Hint: When you apply the ratio test, you are taking the ratio of two numbers. That is only valid if the denominator is nonzero. – 2017-02-07
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    @rogerl But my denominator is going to cancel with terms in my $a_{n+1}$ numerator... I'm not figuring it out – 2017-02-07
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    When you write $\frac{a_{n+1}}{a_n}$, you are already assuming that $a_n\ne 0$, since otherwise what you wrote is $\frac{0}{0}$. – 2017-02-07
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    Ok, so for my denominator $a_n$ I need to assume that $x \neq \frac{1}{2}$... So the ratio test is working for all values except when $x = \frac{1}{2}$. Then I need to check if it's going to converge or not for that value... wich makes my equation $= 0$, and because of that converges. Is that all correct? – 2017-02-07

2 Answers 2

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When $x\neq 1/2$:

$$\lim_{n\to\infty} \left|\frac{t_{n+1}}{t_n} \right|= \lim_{n\to\infty} (n+1) |2x-1| < 1$$

which is never satisfied.

When $x=1/2$ the sum becomes

$$1+0+0+\cdots = 1$$

and converges trivially.

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    But with $x=\frac{1}{2}$ we'll get that limit to be $0\cdot \infty$ and that's a undetermined value. – 2017-02-07
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    @Bruno: No, when $x=\frac12$ the series is $1+0+0+0+0+\cdots$ which converges to $1$. – 2017-02-07
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    @BrunoReis It is not undetermined if the limit is invoked properly. I have updated the answer to demonstrate this. – 2017-02-07
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    @JonasMeyer Thanks. – 2017-02-07
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    Perfect my friend!!! Thanks – 2017-02-08
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Definition: $R\in [0,\infty]$ is the radius of convergence of the series $\sum_na_n(x-A)^n$ iff (1) the series converges whenever $|x-A|

Observe that if $R=0$ then there are no $x$ with $|x-A|

The Hadamard Radius Formula.... Theorem: Let $T=\lim_{n\to \infty}\sup_{m>n}|a_m|^{1/m}.$ The series $\sum_na_n(x-A)^n$ converges when $|x-A|<1/T$ and diverges when $|x-A|>1/T.$ (With the convention, for brevity, that $1/0=\infty$ and $1/\infty=0.$)