Write the Maclaurin series for f(x)=dfrac11+x^k, k in N
Write the Maclaurin series for $f(x)=\dfrac1{1+x^k}$, $k\in\mathbb N$.
Explanation of Method
Substituting $t=-x^k$ into the geometric series $\frac1{1-t}=\sum t^n$ gives the expansion directly; the region of convergence follows from $|t|<1$, and both boundary points diverge regardless of $k$, since the terms there always have magnitude exactly $1$.
- Substitute $t=-x^k$ into the geometric series.
- Determine the region of convergence $|x|<1$.
- Check both endpoints $x=\pm1$: terms never vanish, for any $k$.
Step 1 · Substitution
With $t=-x^k$: $$ \frac1{1+x^k} = \frac1{1-(-x^k)} = \sum_{n=0}^{\infty}(-x^k)^n = \sum_{n=0}^{\infty}(-1)^nx^{kn}. $$
Step 2 · Region of convergence
The geometric series converges iff $|t|<1$: $|{-x^k}|<1\iff|x|^k<1\iff|x|<1$. So $R=1$, independent of $k$.
Step 3 · Endpoints $x=\pm1$
At $x=\pm1$: $|x^k|=1$ for every $k$, so each term $(-1)^nx^{kn}$ has magnitude exactly $1$ and never tends to $0$. By the Divergence Test, the series diverges at both $x=1$ and $x=-1$, regardless of $k$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum(-1)^nx^{kn}$ |
| Radius of convergence | $R=1$, independent of $k$ |
| At $x=\pm1$ | Diverges (terms have magnitude $1$) |
Additional Notes
- Only every $k$-th power of $x$ appears with a nonzero coefficient, giving the series a "sparse" structure directly analogous to Problem 255's $1/(4-x^4)$, which is the special case obtained by rescaling ($4-x^4=4(1-x^4/4)$, matching $k=4$ up to the constant factor).
- For $k=1$, this reduces to the ordinary alternating geometric series for $1/(1+x)$; for $k=2$, term-by-term integration recovers the Maclaurin series for $\arctan x$, exactly the reverse operation used in Problem 239.
- The radius of convergence being exactly $1$ for every natural number $k$ reflects that the nearest singularity of $1/(1+x^k)$ to the origin (a $k$-th root of $-1$) always has modulus $1$, regardless of $k$.