1
$\begingroup$

This question comes to my mind immediately after asking this question.

I was earlier unknown that limit of sum equal sum of limits only when there are finite terms. Now the problem is then how do I evaluate the following limit which earlier I used to do by applying individual limits.

$$ \lim_{x \to 0}\left(-\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots\right) $$

I’m high school student

$\endgroup$
5
  • 1
    $\begingroup$ It looks like $\frac{\sin(x)-x}{x^3}$ $\endgroup$ – Gribouillis Jul 10 '20 at 8:09
  • $\begingroup$ All the terms will tend to zero apart from the first one, so there is your limit $\endgroup$ – Henry Lee Jul 10 '20 at 8:26
  • $\begingroup$ @HenryLee That argument does not work for infinite sums. $\endgroup$ – José Carlos Santos Jul 10 '20 at 9:46
  • $\begingroup$ @JoséCarlosSantos Why will it not? $\endgroup$ – wesupportthepalace Jul 10 '20 at 14:36
  • $\begingroup$ Take, for instance, for each $n\in\Bbb N$ and each $x\in(0,\infty)$,$$f_n(x)=\begin{cases}nx&\text{ if }x\in\left(0,\frac1n\right)\\1&\text{ otherwise.}\end{cases}$$Now, for each $n\in\Bbb N$, let$$g_n=\begin{cases}f_1&\text{ if }n=1\\f_n-f_{n-1}&\text{ otherwise.}\end{cases}$$Then\begin{align}(\forall x\in(0,\infty):\sum_{n=1}^\infty g_n(x)&=f_1(x)+\bigl(f_2(x)-f_1(x)\bigr)+\bigl(f_3(x)-f_2(x)\bigr)+\cdots\\&=\lim_{n\to\infty}f_n(x)\\&=1\end{align}and therefore $\lim_{x\to0}\sum_{n=1}^\infty g_n(x)=1$. However, for each $n\in\Bbb N$, $\lim_{x\to0}g_n(x)=0$. $\endgroup$ – José Carlos Santos Jul 10 '20 at 14:52
7
$\begingroup$

That limit is $-\frac1{3!}$. That's so because, when a power $\sum_{n=0}^\infty a_nx^n$ series has radius of convergence $r$ greater than $0$ (and the radius of convergence of your series is $\infty$), then, if $f(x)=\sum_{n=0}^\infty a_nx^n$ ($|x|<r$), $f$ is a continuous function. In particular,$$a_0=f(0)=\lim_{x\to 0}f(x)=\lim_{x\to0}\sum_{n=0}^\infty a_nx^n.$$


Here is a more elementary approach. For each real number $x$ such that $|x|<1$,\begin{align}\left|\frac{x^2}{5!}-\frac{x^4}{7!}+\cdots\right|&\leqslant\frac{|x|^2}{5!}+\frac{|x|^4}{7!}+\cdots\\&\leqslant\frac{|x|^2}{120}\left(1+|x|^2+|x|^4+\cdots\right)\\&=\frac{|x|^2}{120\left(1-|x|^2\right)}\end{align}And so, since $\lim_{x\to0}\frac{|x|^2}{120\left(1-|x|^2\right)}=0$,$$\lim_{x\to0}-\frac1{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+\cdots=-\frac1{3!}+0=-\frac1{3!}.$$

$\endgroup$
4
  • 2
    $\begingroup$ I’m a high school student and unable to understand your answer $\endgroup$ – dRIFT sPEED Jul 10 '20 at 8:19
  • $\begingroup$ @pRSmHJN1 Executive summary: as long as $f(x)=a_0+a_1x+a_2x^2+\cdots$ converges somewhere away from zero, then $\lim_{x\to0}f(x)=a_0$. $\endgroup$ – Angina Seng Jul 10 '20 at 8:27
  • $\begingroup$ @pRSmHJN1 I have added a more elementary answer. $\endgroup$ – José Carlos Santos Jul 10 '20 at 8:28
  • 1
    $\begingroup$ +1 for the elementary proof. Either one has to use ideas of power series, uniform convergence, or settle down for a elementary proof suited for specific series, but in no case one should be made to believe that the result is automatic. $\endgroup$ – Paramanand Singh Jul 10 '20 at 12:22
2
$\begingroup$

Observe that, for $x\ne 0$, $$ -\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots=\frac{1}{x^3} \left(-\frac{x^3}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9!}+\cdots\right) =\frac{1}{x^3} \left(\frac{x}{1!}-\frac{x^3}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9!}+\cdots-\frac{x}{1!}\right) = \frac{1}{x^3}\left(\sin x-x\right) \\ $$ Then you may apply L'Hôpital three times and obtain that $$ \lim_{x\to 0}\frac{1}{x^3}\left(\sin x-x\right)=-\frac{1}{3!} $$

$\endgroup$

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.