Find the domain of convergence and the sums
Find the domain of convergence and the sums $f(r,\theta)=\sum_{n=1}^{\infty}r^n\cos(n\theta)$ and $\varphi(r,\theta)=\sum_{n=1}^{\infty}r^n\sin(n\theta)$.
Explanation of Method
Both series are the real and imaginary parts of the complex geometric series $\sum_{n=1}^{\infty}(re^{i\theta})^n$, which converges for $|r|<1$ regardless of $\theta$. Summing the geometric series in closed form and rationalizing the denominator separates the real and imaginary parts directly.
- Combine the two series into a single complex geometric series $\sum(re^{i\theta})^n$.
- Sum the geometric series and rationalize to isolate real and imaginary parts.
- Verify both closed forms numerically against direct term-by-term summation.
Step 1 · Domain of convergence
Since $|r^ne^{in\theta}|=|r|^n$, the complex series $\sum_{n=1}^{\infty}(re^{i\theta})^n$ converges absolutely iff $|r|<1$, for every $\theta$. Both $f$ and $\varphi$ therefore converge exactly on $$ \boxed{|r|<1.} $$
Step 2 · Sum the complex geometric series
$$ \sum_{n=1}^{\infty}(re^{i\theta})^n = \frac{re^{i\theta}}{1-re^{i\theta}}. $$ Rationalizing by multiplying numerator and denominator by $1-re^{-i\theta}$: $$ \frac{re^{i\theta}(1-re^{-i\theta})}{(1-re^{i\theta})(1-re^{-i\theta})} = \frac{re^{i\theta}-r^2}{1-2r\cos\theta+r^2} = \frac{(r\cos\theta-r^2)+ir\sin\theta}{1-2r\cos\theta+r^2}. $$
Step 3 · Separate real and imaginary parts
Matching real and imaginary parts against $f+i\varphi$: $$ f(r,\theta) = \frac{r\cos\theta-r^2}{1-2r\cos\theta+r^2} = \frac{r(\cos\theta-r)}{1-2r\cos\theta+r^2}, \qquad \varphi(r,\theta) = \frac{r\sin\theta}{1-2r\cos\theta+r^2}. $$ (Verified numerically at $r=0.3,\theta=0.7$: a 300-term direct sum gives $f\approx0.220969$ and $\varphi\approx0.306238$, matching both closed forms to 6 decimal places. Note the real-part formula must have the factor $-r^2$, not merely $1$, in its numerator — the more familiar-looking $\dfrac{1-r\cos\theta}{1-2r\cos\theta+r^2}$ is the sum starting from $n=0$ instead of $n=1$, and differs from $f$ by exactly $1$.)
Summary Table
| Property | Value |
|---|---|
| Domain | $|r|<1$ (all $\theta$) |
| $f(r,\theta)=\sum r^n\cos n\theta$ | $\dfrac{r(\cos\theta-r)}{1-2r\cos\theta+r^2}$ |
| $\varphi(r,\theta)=\sum r^n\sin n\theta$ | $\dfrac{r\sin\theta}{1-2r\cos\theta+r^2}$ |
Additional Notes
- The denominator $1-2r\cos\theta+r^2$ is the classical Poisson-kernel denominator, also the origin of the Chebyshev-generating-function identities used in Problems 261–263: $|1-re^{i\theta}|^2=1-2r\cos\theta+r^2$.
- Adding $1$ to $f(r,\theta)$ recovers the $n=0$-inclusive sum $\dfrac{1-r\cos\theta}{1-2r\cos\theta+r^2}$, the standard real Poisson kernel used in harmonic-function boundary-value problems — the distinction between the two forms is exactly analogous to the missing-constant pitfall in Problem 274.
- Setting $\theta=0$ collapses both formulas to the real-axis check: $f(r,0)=\dfrac{r(1-r)}{(1-r)^2}=\dfrac r{1-r}$ (the ordinary geometric series $\sum r^n$) and $\varphi(r,0)=0$, as expected.