I am trying to solve the following question:
Prove that the series $$\sum_{n = 1}^{ \infty} \frac{x^{2n – 1}}{(2n – 1)!} $$ and the series $$\sum_{n = 1}^{ \infty} \frac{x^{2n}}{(2n)!}$$ both converges uniformly on any bounded interval of $\mathbb R.$
I was thinking about using the Weierstrass M – test, but this test does not speak about any bounded interval.
Here is the statement of the Weierstrass M – test:
Let $\{f_n\}$ be a sequence of functions defined on a common domain $A,$ and let $\{M_n\}$ be a sequence of positive numbers such that $|f_n(x)| \leq M_n,$ for each $ n \in \mathbb N,$ and any $x \in A.$
If the series $\sum_{n = 1}^{\infty} M_n$ converges, then $\sum_{n = 1}^{\infty} f_n$ converges uniformly.
Could someone explain to me how I can in general tackle this kind of problems please?
Best Answer
You say the Weierstrass test doesn't talk about bounded intervals, and you are right. But, what it does talk about is uniform convergence on a given set $A$; if $\sum\limits_{n=1}^{\infty}\sup\limits_{x\in A}|f_n(x)|<\infty$, then the series converges uniformly on $A$ (and recall that uniform convergence on $A$ implies uniform convergence on every subset of $A$). In other words, take $M_n=\sup\limits_{x\in A}|f_n(x)|$.
Let's look at the series $\sum\limits_{n=1}^{\infty}\frac{x^{2n}}{(2n)!}$. Here, we have $f_n:\Bbb{R}\to\Bbb{R}$, $f_n(x)=\frac{x^{2n}}{(2n)!}$. Suppose $A$ is any bounded interval. This means there exists some $r>0$ such that $A\subset [-r,r]$. We have \begin{align} \sum_{n=1}^{\infty}\sup_{x\in A}\left|\frac{x^{2n}}{(2n)!}\right|\leq \sum_{n=1}^{\infty}\sup_{x\in [-r,r]}\left|\frac{x^{2n}}{(2n)!}\right| \leq \sum_{n=1}^{\infty}\frac{r^{2n}}{(2n)!}<\infty. \end{align} The first estimate should be obvious, since $A$ is contained in $[-r,r]$; this is just a basic property of supremums and set inclusions. The second inequality is actually an equality in this special case; I left it as a weak inequality just to emphasize that we don't have to be (and shouldn't aim to be unless explicitly instructed so) super precise with the estimates. All we want to do is show a certain numerical series is finite. The final step of claiming the series is finite can be done for example using the ratio test. In fact the final series equals $\cosh(r)-1$.
All the hypotheses of Weierstrass' M-test are satisfied, hence the series converges uniformly on $A$. Since $A$ was taken to be an arbitrary bounded subset of $\Bbb{R}$, this proves that the series converges uniformly on every bounded subset of $\Bbb{R}$.
In about 95% of situations, you can prove uniform convergence simply by the Weierstrass M-test (very rarely have I had to use some other method, such as Dirichlet's test; in fact I haven't used it recently, so much so that I kind of even lose track of the precise assumptions).
The Weierstrass M-test deals with uniform convergence of arbitrary functions. Specifically for power series, there is a very slight improvement. The essence is still Weierstrass' test. Anyway, the statement is:
We assume $z_0\neq 0$ because the series always converges at the origin; so we're just excluding the trivial case. The "strength" of this theorem is that our hypothesis only tells us the series $\sum_{n=0}^{\infty}a_nz_0^n$ converges; we don't know anything about absolute convergence.
TO prove this, note that since the series converges, the general summand must tend to zero: $a_nz_0^n\to 0$ as $n\to\infty$. In particular, it is a bounded sequence; i.e $M:=\sup\limits_{n\geq 1}|a_nz_0^n|<\infty$. Now, for any $0\leq r<|z_0|$, we have \begin{align} \sum_{n=0}^{\infty}\sup_{|z|\leq r}\left|a_nz^n\right|=\sum_{n=0}^{\infty}\sup_{|z|\leq r}\left|\frac{z^n}{z_0^n}a_nz_0^n\right|\leq\sum_{n=0}^{\infty}\frac{r^n}{|z_0|^n}M<\infty, \end{align} by the ratio test with the common ratio $\rho=\frac{r}{|z_0|}<1$ (actually, you can explicitly sum the geometric series; the answer is $\frac{M}{1-(r/|z_0|)}<\infty$). By Weierstrass' test, it follows the series converges uniformly on the closed disk $D_r=\{z\in\Bbb{C}\,:\, |z|\leq r\}$, hence the proof is complete.