Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{x^n}{1+x^{2n}}. $$
Explanation of Method
The behavior splits naturally at $|x|=1$: for $|x|<1$ the denominator's $x^{2n}$ term is negligible, so the series behaves like the geometric series $\sum x^n$; for $|x|>1$, factoring $x^{2n}$ out of the denominator instead makes it behave like $\sum1/x^n$. At $|x|=1$ exactly, every term equals a nonzero constant, so the series diverges outright.
- Compare the term for $|x|<1$ against $|x|^n$.
- Compare the term for $|x|>1$ against $|x|^{-n}$.
- Check $x=\pm1$ directly.
Step 1 · Case $|x|<1$
$\left|\dfrac{x^n}{1+x^{2n}}\right|\le|x|^n$ (since $1+x^{2n}\ge1$), and $\sum|x|^n$ converges for $|x|<1$, so by comparison the series converges absolutely.
Step 2 · Case $|x|>1$
Factoring: $\dfrac{x^n}{1+x^{2n}}=\dfrac1{x^n+x^{-n}}$, and $\left|\dfrac1{x^n+x^{-n}}\right|<\dfrac1{|x|^n}$ (dropping the positive term $x^{-n}$ from the denominator only increases the bound, in absolute value terms). Since $\sum1/|x|^n$ converges for $|x|>1$, the series converges absolutely.
Step 3 · Endpoints $x=\pm1$
At $x=1$: term $=\dfrac1{1+1}=\dfrac12$ for every $n$, so $\sum\frac12$ diverges. At $x=-1$: term $=\dfrac{(-1)^n}2$, which doesn't tend to $0$, so the series diverges by the $n$-th term test.
Summary Table
| Region | Behavior |
|---|---|
| $|x|\ne1$ | Absolutely convergent |
| $x=\pm1$ | Diverges |
Additional Notes
- This series' two-sided geometric-type behavior (small $|x|$ like $\sum x^n$, large $|x|$ like $\sum x^{-n}$) mirrors the complementary comparison arguments of Problems 164 and 165, both handled via a single-direction geometric bound rather than this two-sided split.
- The exact cancellation at $|x|=1$ — where the term collapses to a constant $\frac12$ rather than vanishing — is the key qualitative difference from a typical power series boundary, where terms usually don't stabilize at a nonzero constant so cleanly.
- Since the series converges absolutely on both sides of $|x|=1$ with no boundary subtlety beyond outright divergence, there is no region of conditional convergence anywhere.
Result
Converges absolutely for $|x|\ne1$, diverges at $x=\pm1$: