Been revising in advance for complex analysis, and struggling with questions relating to power series. I've got the following series,

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

and I'm asked to find values of $x$ where $g(x)$ converges, and to show it's continuous there. I tried doing a ratio test to get an interval of convergence, and somehow got $\frac{1}{\sqrt{x}}$ which probably isn't right, since I'm fairly certain this converges for values $x \in [0,1]$ although I'm unsure as to how to explain the reasoning there. I also know I should use the Weierstrass M-Test.... but I'm not sure how to "cook up" an appropriate convergent series.

Given that though, continuity should follow from term-by-term continuity on the defined interval, right?


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    $\begingroup$ Try the root test. It might tell you something. $\endgroup$ Jul 20 '17 at 2:14
  • $\begingroup$ This is not a power series, so radius of convergence and the like is not really correct terminology. $\endgroup$
    – zhw.
    Jul 20 '17 at 2:25
  • $\begingroup$ Thought had crossed my mind but thanks, edited.. Somehow we never covered the root test @ncmathsadist in either of the analysis courses I've taken, and embarrassingly I still don't know how to calculate a limit superior... $\endgroup$ Jul 20 '17 at 2:27

Note that if $|x|\ge1$, then $g(x)\ge \sum_{n=0}^\infty\frac{x^{2n}}{2x^{2n}}=\sum_{n=0}^\infty \frac{1}{2}$ diverges to infinity.

You can show that $g$ converges pointwise in $(-1,1)$ by showing that $g$ converges uniformly on every compact subinterval $[-\beta,\beta]\subset(-1,1)$. For the denominator satisfies $1+x^{2n}\ge 1$ and on $[-\beta,\beta]$ the numerator satisfies $x^{2n}\le \beta^{2n}$. Thus $$ \frac{x^{2n}}{1+x^{2n}}\le\beta^{2n}. $$ As $\beta<1$, can you apply the $M$-test to finish? Moreover as the partial sums are continuous on $[-\beta,\beta]$, can you conclude that $g$ is continuous on $[-\beta,\beta]$ and hence continuous on all of $(-1,1)$?

  • $\begingroup$ Oh cool that all makes sense, except for why is $x^{2n} \leq \beta^{2n}$? $\endgroup$ Jul 20 '17 at 2:31
  • $\begingroup$ @AdrianHindes Since $|x| \le \beta$ for every $x\in [-\beta,\beta]$, if we square both sides of this inequality, we get the answer to your question. $\endgroup$
    – Alex Ortiz
    Jul 20 '17 at 2:38
  • $\begingroup$ Oh right sorry, I thought for a moment we were taking $x$ in a larger interval. Okay great that makes perfect sense then, thanks so much! $\endgroup$ Jul 20 '17 at 2:41

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