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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?

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    $\begingroup$ @GEdgar so using this test is the general way of solving problems like this? if so, how can we guess the sequence $\{M_n\}$? $\endgroup$
    – Brain
    Jan 11 at 2:06
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    $\begingroup$ @Brain Well, take the obvious candidates: $$ \frac{{K^{2n - 1} }}{{(2n - 1)!}}, \quad \frac{{K^{2n} }}{{(2n)!}}. $$ $\endgroup$
    – Gary
    Jan 11 at 2:15
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    $\begingroup$ You can find the radius of convergence $R$ of each series. It will be $\infty$. That implies that each series is converging uniformly on every closed interval. $\endgroup$
    – markvs
    Jan 11 at 2:17
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    $\begingroup$ I remember this theorem from the second semester analysis course I took 45 years ago. It was in a different country and different language. I am sure there is an English version of this result and there are analysis specialists here who can give a reference. I can reproduce a proof which was easy but a reference would be better. $\endgroup$
    – markvs
    Jan 11 at 2:46
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    $\begingroup$ @Brian, yes, Abel's theorem: ib.mazurok.com/2015/06/04/… $\endgroup$
    – markvs
    Jan 12 at 1:18

3 Answers 3

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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:

Let $\{a_n\}_{n=0}^{\infty}$ be a sequence of complex numbers and $z_0$ a non-zero complex number such that the series $\sum_{n=0}a_nz_0^n$ converges. Then, for any $0\leq r<|z_0|$ (note the strict inequality), the series $\sum_{n=0}^{\infty}a_nz^n$ converges absolutely and uniformly on the closed disk $D_r=\{z\in\Bbb{C}\,:\, |z|\leq r\}$.

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.

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  • $\begingroup$ Amazing answer ..... I am going to read it thoroughly. $\endgroup$
    – Brain
    Jan 11 at 3:27
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Those are power series of a single (might as well let it be complex) variable. If $R$ is the radius of convergence of the power series, then for every $R' < R$, the power series converges absolutely and uniformly on the closed disk $\overline{D_{R'}(0)} = \{z \in \mathbb{C} : |z| \leq R'\}$. The classical proof for this can be found on page 102 here: https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/anal1v.pdf

An somewhat analogous result holds for power series of several complex variables, and the proof is not very different. Proofs for this more general situation can be found on page 58 here: https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/analmv.pdf

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  • $\begingroup$ You may write $z\in \mathbb C$ in your first paragraph as you are talking about one complex variable. $\endgroup$
    – Gary
    Jan 11 at 3:16
  • $\begingroup$ @Gary Yeah thanks that's what I meant. $\endgroup$
    – Mason
    Jan 11 at 3:19
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You can solve it using the M-test too. For the first series Let $[a,b] \subset \mathbb R$ and let for each $n \in \mathbb N$ denote $B_n = \sup_{x \in [a,b] } |x|^{2n-1} < \infty$. Let $f_n(x) = \frac{x^{2n -1}}{(2n-1)!}$ then choose $N>0$ large enough such that $B_n < e^n$, then we have

$$|f_n(x)| \leq \frac{B_n}{(2n-1)!} \leq \frac{e^n}{(2n-1)!}= M_n$$

Since $\sum_{n=1}^\infty M_n = \sqrt{e} \sinh(\sqrt{e}) \approx 4.1284 $ (converges). Therefore, the main series is uniformly convergent. You can argue the same way for the second series.

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