1
$\begingroup$

I'm trying to show uniform convergence of the power series of $\sin x$ on $[0,r] (r>0)$ from first principles. So let $\epsilon>0$ and $n\geq N$. I'm stuck at this step: What do I choose for $N$ (a function of $\epsilon$) so that

$$\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!} <\epsilon ?$$

(I know the usual method is to observe that the LHS expression above converges to $0$, but here I would like to use a pure $N-\epsilon$ argument.)

$\endgroup$
0
2
$\begingroup$

Using part of Stirling's approximation implying that $k!>\left(\frac{k}{e}\right)^k$ allows us to get a crude geometric series bound when $n$ is big enough:

$$ \begin{align*} \sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!} &=r\sum\limits_{k=n+1}^\infty\frac{(r^2)^k}{(2k+1)!}\\ &< r\sum\limits_{k=n+1}^\infty\frac{(r^2)^k}{k!}\\ &<r\sum\limits_{k=n+1}^\infty\left(\dfrac{er^2}{k}\right)^k\\ &<r\sum\limits_{k=n+1}^\infty\left(\dfrac{er^2}{n+1}\right)^k.\\ \end{align*} $$

The geometric series in the last expression could diverge to $+\infty$. However, it is always okay to add assumptions that require $n$ to be larger. In particular, if $n+1>er^2$, then the geometric series converges, showing that $$\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!}<\dfrac{r\left(\frac{er^2}{n+1}\right)^{n+1}}{1-\frac{er^2}{n+1}}.$$

In order to get a bound that is easier to work with, we can use the inequality $\dfrac{a^k}{1-a}<2^{k-1}$ if $0<a<\frac12$. So making the stronger assumption that $n+1>2er^2$ implies that $$\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!}<r2^{-n}.$$

We easily see that $n>\dfrac{\ln(r/\varepsilon)}{\ln(2)}$ implies that $r2^{-n}<\varepsilon$, so under the standing assumption that $n+1>2er^2$, the inequality $n>\dfrac{\ln(r/\varepsilon)}{\ln(2)}$ implies that $\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!} <\epsilon$.

Thus $N>\max\left\{\dfrac{\ln(r/\varepsilon)}{\ln(2)},2er^2\right\}$ suffices, because any $n>N$ automatically satisfies $n+1>2er^2$, so that the bound $r2^{-n}$ for our series applies.

$\endgroup$
5
  • $\begingroup$ Isn't "$n+1>2er^2$" already covered by the case before it, and what about the case whereby $n+1\leq er^2$? And how does the final line lead to the desired conclusion (I can only see that $\dfrac{\ln(r/\varepsilon)}{\ln(2)}$ corresponds to desired conclusion for the second case right above it)? $\endgroup$
    – ryang
    Sep 24 '13 at 6:47
  • $\begingroup$ @Ryan: No, $n+1>er^2$ does not imply that $n+1>2er^2$, but conversely. The point of $n+1>er^2$ is to make the bound preceding it finite (a convergent geometric series). The point of adding $n+1>2er^2$ is to make the expression $\dfrac{er^2}{n+1}$ less than $1/2$, which gives the upper bound of $r2^{-n}$. Note that $N>\frac{\ln(r/\varepsilon}{\ln(2)}$ corresponds to ensuring that $n\geq N$ implies $r2^{-n}<\varepsilon$. And if $N>2er^2$, then $n\geq N$ implies $\sum_{k=n+1}^\infty \frac{r^{2k+1}}{(2k+1)!}<r2^{-n}$, as shown above. Thus the max is taken to be able to combine the inequalities. $\endgroup$ Sep 24 '13 at 12:57
  • $\begingroup$ @Ryan: I forgot to answer "what about...$n+1\leq er^2$"? There is no need to consider that case. We want $N$ such that $n\geq N$ implies something is less than $\varepsilon$. If one $N$ works, any larger $N$ will work. It was convenient for the method I used to take $N>er^2$, and even $N>2er^2$. We need not worrying about the finitely many $n$ that come before this. $\endgroup$ Sep 24 '13 at 13:13
  • $\begingroup$ Ah, thank you so much for your answer, there is no way I could've written out this $N-\epsilon$ proof on my own. Please consider removing the $n+1>er^2$ case because its inclusion is unnecessary and frustratingly confusing, and led me to think that you were taking cases and also led me to futilely try to fit the $2er^2$ term inside the max to that case, since the other term inside the max clearly corresponds to the other case, and this in turn made me lose track of how the final line led to the desired conclusion. (I do understand it all now, thanks to the clarifications.) $\endgroup$
    – ryang
    Sep 25 '13 at 7:55
  • $\begingroup$ @Ryan: I will not completely remove that, because the point is to use that $\sum_{n=k}^\infty a^n=\dfrac{a^k}{1-a}$ if $|a|<1$. Subsequently the stronger assumption is added in order to make $\dfrac{a^k}{1-a}$ bounded with something simple. (All of these assumptions point in the direction of making $N$ bigger, which is fine, just like in an $\varepsilon$-$\delta$ continuity proof it is always okay to make assumptions that make $\delta$ smaller, and take a $\min$). However, I may at some point find time to rewrite for better clarity. I do not want it to be confusing; thank you for the feedback. $\endgroup$ Sep 25 '13 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.