Find the values of $x$ where $\sum_n \frac{x^{2n}}{1+x^{2n}}$ converges, and show that we have a continuous function on this set

I can see that

$$\frac{x^{2n}}{1+x^{2n}}<x^{2n},\forall x\in(-1,1),\forall n\in\Bbb N$$

so I can deduce that at least the series converges in $(-1,1)$, and if we take $|x|=1$ the series diverges, then the series converges only in the interval $(-1,1)$.

Now my problem is that Im unable to show that the limit function is continuous. A theorem says that if the series converges uniformly and each $f_n$ is continuous then $f$ is continuous as well.

Another theorem that I know about the continuity of the limit function of a series is: if $f_n$ is differentiable and $(f'_n)\to g$ uniformly, and exist some $x_0$ where $f_n$ converges then $(f_n)\to f$ uniformly and $f'=g$. But I dont get something useful from here either.

But Im unable to show that $\sum_n \frac{x^{2n}}{1+x^{2n}}$ converges uniformly on $(-1,1)$, I tried some M-test but I cant get something useful, can you help me with this question?

  • $\begingroup$ suppose that $\sum_{n \ge 0} |a_n|$ converges, can you prove that $f(x) = \sum_{n \ge 0} a_n x^n$ is continuous on $|x| < 1$ ? $\endgroup$ – reuns Jun 16 '16 at 0:38
  • $\begingroup$ note that if $\sum_{n \ge 0} |a_n|$ converges, then obviously $|a_n| < C$ so that (for $|x| < 1$) : $\sum_{n \ge 0} a_n x^n < C \sum_{n \ge 0} x^n = \frac{C}{1-x}$, this is all you need for proving that power series are $C^\infty$ inside their disk of convergence $\endgroup$ – reuns Jun 16 '16 at 0:44
  • $\begingroup$ @user1952009 This is not expressed as a power series, however. $\endgroup$ – Clement C. Jun 16 '16 at 1:10
  • $\begingroup$ @ClementC. : it changed, it was $\sum_n \frac{x^{2n}}{1+2^{2n}}$ $\endgroup$ – reuns Jun 16 '16 at 1:11
  • $\begingroup$ I missed that. In context, your comment makes much more sense then :) $\endgroup$ – Clement C. Jun 16 '16 at 1:12

tl;dr: fix any $a \in (0,1)$ , and show normal convergence of the series $\sum_n f_n$ on $[-a,a]$. This will imply continuity of $f=\sum_n f_n$ on $[-a,a]$; since this holds for any $a\in(0,1)$, $f$ will be continuous on $(-1,1)$.

First detail: You didn't actually proved convergence only on $(-1,1)$, since your argument uses one that usually applies to power series (radius of convergence). But it's true nevertheless: letting $f_n(x) \stackrel{\rm def}{=} \frac{x^{2n}}{1+x^{2n}}$, we first restrict to $x\geq 0$ (as all $f_n$'s are even), and get:

  • if $0\leq x < 1$, $f_n(x) \sim_{n\to\infty} x^{2n}$ and the series converges by comparison;
  • if $x=1$, $f_n(1) = \frac{1}{2}$ and the series trivially diverges;
  • if $x>1$, then $f_n(x) \sim_{n\to\infty} 1$ and the series trivially diverges by comparison.

so indeed you only have convergence on $I\stackrel{\rm def}{=}(-1,1)$.

Now, continuity: To prove continuity of $f=\sum_n f_n$ on $I$: since continuity is a local property, it suffices to show $f$ is continuous on every $[-a,a]\subseteq I$. Which is much easier... since you have normal convergence on every such $[-a,a]$.

That is, fix $a \in [0,1)$, and show the series $$ \sum_{n=1}^\infty \sup_{x\in[-a,a]} \lvert f_n(x)\rvert $$ converges.$^{(\dagger)}$ This is normal convergence of the series on $[-a,a]$, which implies uniform convergence of the series on $[-a,a]$, which implies continuity of $f$ on $[-a,a]$ since the $f_n$'s are continuous. Since $a$ was arbitrary, $f$ is continuous on $(-1,1)$.

$(\dagger)$ This is immediate, as each $f_n$ is even and increasing on $[0,a]$: $$ \sup_{x\in[-a,a]} \lvert f_n(x)\rvert = \lvert f_n(a)\rvert = \frac{a^{2n}}{1+a^{2n}} $$ and since $a\in(0,1)$...

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Note that local uniform convergence preserves continuity, hence, it suffices to prove that $(S_n)$ converges to $S$ on a neighbourhood of each point in $(-1,1)$, for which it suffices to prove convergence over intervals of the form $[-a,a]$ with $0<a<1$, where:

$$S_n(x) = \sum_{k=0}^n \frac{x^{2k}}{1 + x^{2k}}$$


$$S(x) = \sum_{k=0}^{\infty} \frac{x^{2k}}{1 + x^{2k}}$$

Let $0 < a <1$. Then,

$$|S_n(x)- S(x)| = \left| \sum_{k=n+1}^{\infty}\frac{x^{2k}}{1 + x^{2k}} \right| \le \left|\sum_{k=n+1}^{\infty} x^{2k} \right| \le \sum_{k=n+1}^{\infty} a^{2k}$$

for all $x \in [-a,a]$. So,

$$\|S_n - S \|_{\infty} \le \sum_{k=n+1}^{\infty} a^{2k}$$

The RHS converges to $0$ since the series $\sum a^{2n}$ is convergent. So we get the result.

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The radius of convergence is 2.

$\frac {x^{2n}}{1+2^{2n}} <(\frac {x}{2})^{2n}$

If $\sum (\frac {x}{2})^{2n}$ converges $\sum \frac {x^{2n}}{1+2^{2n}}$ by the comparison test.

There are 4 tests to know. The direct comparison test (as illustrated above) the limit comparion test. if $|\lim_\limits{n\to \infty} \frac {a_n}{b_n}| = M$ with $0<M<\infty$. Either both series converge or both series diverge.

The ratio test: if $\lim_\limits{n\to \infty}|\frac {a_{n+1}}{a_n} x| < 1$ the series converges. $r >\lim_\limits{n\to \infty}|\frac {a_{n}}{a_n+1}|$ is the radius of convergence

the root test: if $\lim_\limits{n\to \infty}(|a_{n}|)^{\frac 1n} |x| < 1$ the series converges. $r >\lim_\limits{n\to \infty}(|a_{n}|)^{\frac 1n}$ is the same radius of convergence.

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  • 1
    $\begingroup$ Sorry, I had a typo in the title, the function actually is $\sum x^{2n}/(1+x^{2n})$ $\endgroup$ – Masacroso Jun 16 '16 at 0:46
  • $\begingroup$ @Masacroso : note that it doesn't change fundamentally the argument. most people think to $\sum_n \frac{x^{2n}}{1+x^{2n}}$ as : for $|x| < 1-\epsilon$ each term in absolute value is $< \frac{x^{2n}}{\epsilon}$ and $\sum_n \frac{x^{2n}}{\epsilon} = \frac{1}{\epsilon} \frac{1}{1-x^2}$ $\endgroup$ – reuns Jun 16 '16 at 1:13

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