# Divergence of power series

In my lectures we considered a power series and used the ratio test for absolute convergence to find the radius or interval of convergence.

However it was also stated in my lectures that "we have only proven that the series does not converge absolutely for $|x|>1/L$ but a stronger statement that the series diverges here can be proven."

Clearly if the series converges absolutely, it will generally converge. However it is not at all obvious to me as to why a power series which has $|x|>1/L$ necessarily diverges for both positive and negative x, or for coefficients of changing sign (in general for a non-positive series). I have not been able to find much about this online.

So my questions are:

• why is the above true?

• is there any significance of this at the radius of convergence/ interval endpoints? For example, at the radius I know the ratio test is indeterminate so the series may show either absolute convergence or divergence. But perhaps at the radius of convergence all non-positive, or at least all alternating, series necessarily do converge?

The radius $R$ of convergence for $\sum a_nx^n$ is given by $$\tag1 \frac 1R=\limsup\sqrt[n]{|a_n|}.$$ (If the ratio test works, it will get the same result, but the above is the most general).
Claim. If $|x|>R$, then the series $\sum a_nx^n$ diverges.
Proof. As $\frac1{|x|}<\frac1 R$, there are infinitely many $n$ with $\sqrt[n]{|a_n|}>\frac1{|x|}$. But then $|a_nx^n|>1$ infinitely many often and the series cannot converge. $\square$
While $\sum \frac 1nx^n$ (with $R=1$) converges at $x=-R$ (where it becomes a well-known alternating series) and divereges for $x=R$ (where it becomes the harmonic series), it is possible that for $|x|=R$ the series diverges, even if the series is alternating. Example: $\sum_n (-1)^nx^n$ has $R=1$ and at $x=R$ becomes the divergent altenating series $\sum_n (-1)^n$. Also note that we may "sprinkle" signs ad libitum, i.e., consider a series like $\sum \epsilon_n\frac1nx^n$ with $\epsilon_n\in\{\pm1\}$, whch does not incfluence $R$, but allows us a lot of freedom regarding the convergence and limit or divergence at $x=R$.