November 17th
Today I learned the definition of the limit superior to prove the radius of convergence for a power series. In symbols, we have\[\limsup_{n\to\infty}a_n=\lim_{n\to\infty}\left(\sup_{m \gt n}a_m\right).\]This always exists, provided we allow $\pm\infty.$ If $a_\bullet\to\infty,$ then this comes out to $\infty.$ Otherwise, the sequence $\{a_\bullet\}$ has a maximum, so $\sup_{m \gt \bullet}a_m$ is monotonically decreasing. If the supremum sequence goes to $-\infty,$ then the limit is $-\infty.$ Else, the sequence is lower-bounded, so it has a limit because we just said it's monotonically decreasing.
For the radius of convergence, fix a series $S=\sum_na_n.$ Then, with the root test, examine\[A=\limsup_{n\to\infty}\sqrt[n]{\left|a_n\right|}.\]Note that $L\ge0.$ We have the following cases for the root test; I give the arguments here because I had to reconstruct them today.
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If $A \gt 1$ (including $\infty$), then the sequence $\{a_n\}$ must have infinitely many terms with absolute value at least $1.$ Indeed, if there were only finitely many, then a bound $N$ would exist with $\sup_{m \gt n}\sqrt[m]{|a_m|}\le1$ for $n \gt N,$ implying $A\le1.$
But if an infinite subsequence is bounded below by $1,$ then it diverges by the Divergence test.
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If $A \lt 1,$ then there is a bound $N$ for which $n \gt N$ implies $\sup_{m \gt n}\sqrt[m]{|a_m|} \lt 1-\varepsilon$ for some $\varepsilon \gt 0.$ However, this requires $|a_n| \lt (1-\varepsilon)^n,$ so we get that \[S=\sum_{n=0}^\infty a_n\le\sum_{n=0}^N|a_n|+\sum_{n=N+1}^\infty|a_n| \lt \sum_{n=0}^Na_n+\sum_{n=N+1}^\infty(1-\varepsilon)^n \lt \infty.\] So the series converges, absolutely.
This is relevant to the radius of convergence because it tells us that $a(x)=\sum_na_nx^n$ will converge if $Ax \lt 1,$ and diverge if $Ax \gt 1.$ In other words, we get to explicitly compute the radius of convergence as $\frac1A,$ where $\frac1\infty=0$ and $\frac10=\infty.$
As an aside, this lets us prove that the set of formal power series in $\RR[[x]]$ with positive radius of convergence is a ring. (This can be done with notions of absolute convergence, but whatever.) Of course, the $0$ function and the $1$ function have infinite radius of convergence, so they're safe. For the remaining checks, fix $a(x)=\sum_na_nx^n$ and $b(x)=\sum_na_nx^n$ power series with positive radius convergence. Note that this condition is equivalent to\[A=\limsup_{n\to\infty}\sqrt[n]{|a_n|} \lt \infty,\]and analogously $B \lt \infty$ for $b_n.$ We see that $-a(x)$ also has positive radius of convergence because $|-a_n|=|a_n|,$ so the limit for $-a(x)$ is the same as $a(x)$ and therefore finite.
We construct $a(x)+b(x)$ by adding termwise. Without loss of generality, let's say $A\ge B.$ Then we're interested in\[\limsup_{n\to\infty}\sqrt[n]{|a_n|+|b_n|}\le\limsup_{n\to\infty}\sqrt[n]{2\max\{|a_n|,|b_n|\}}.\]The $\sqrt[n]2$ vanishes as $n\to\infty.$ To be formal with this, we could plug into the definition of $\limsup$ and then expand out the limit of the sequences, but it doesn't look fun. So we're interested in\[\limsup_{n\to\infty}\max\left\{\sqrt[n]{|a_n|},\sqrt[n]{|b_n|}\right\}.\]But this is just $A.$ Explicitly, $A\ge B$ implies that there is some bound $N$ for which $n \gt N$ implies $\sup_{m \gt n}\sqrt[m]{a_m}\ge\frac{A+B}2$ and $\sup_{m \gt n}\sqrt[m]{b_m}\le\frac{A+B}2.$ This means that the maximum always go to $|a_n|$ in the above when past $N,$ making the limit exactly $A.$ It follows that the series has positive radius of convergence again.
We do $a(x)b(x)$ in a similar manner. We're interested in\[\limsup_{n\to\infty}\sqrt[n]{\left|\sum_{k=0}^na_kb_{n-k}\right|}.\]Bounding in the stupidest way possible, this is bounded above by\[\limsup_{n\to\infty}\sqrt[n]{(n+1)\cdot\max_{k\le n}|a_k|\cdot\max_{k\le n}|b_k|}\]Again, the $\sqrt[n]{n+1}$ vanishes, which we don't talk about rigorously. The part to worry about is\[\limsup_{n\to\infty}\sqrt[n]{\max_{k\le n}|a_k|\cdot\max_{k\le n}|b_k|}.\]Expanding the $\limsup$ and doing some rearranging, this collapses to\[\lim_{n\to\infty}\sup_{m \gt n}\max_{k\le m}\sqrt[m]{|a_m|}\cdot\sup_{m \gt n}\max_{k\le m}\sqrt[m]{|b_m|}=\lim_{n\to\infty}\sup_{m\ge0}\sqrt[m]{|a_m|}\cdot\sup_{m\ge0}\sqrt[m]{|b_m|}.\]Limit properties says that this limit is equal to the product of each individual one provided each factor exists, and they do—the fact $A,B \lt \infty$ implies that $\sqrt[n]{a_n}$ and $\sqrt[n]{b_n}$ are bounded above, and therefore they each have a supremum. This finishes.