Evaluate the sums for the following series, for |q|<1
Evaluate the sums for the following series, for $|q|<1$: $$ 1^{\circ}\; S_1 = \sum_{n=1}^{\infty} q^n \sin(nx); \qquad 2^{\circ}\; S_2 = \sum_{n=1}^{\infty} q^n \cos(nx). $$
Explanation of Method
Rather than solving these two real trigonometric series separately, Euler's formula lets them be combined into a single complex geometric series $Z=\sum q^n e^{inx}$, whose real and imaginary parts are exactly $S_2$ and $S_1$. Since $Z$ is an ordinary geometric series in the complex ratio $qe^{ix}$ with $|qe^{ix}|=|q|<1$, it sums to a simple closed form; rationalizing that complex fraction then separates it back into real and imaginary pieces.
- Combine $S_1$ and $S_2$ into a single complex series $Z = S_2 + iS_1$ via Euler's formula.
- Sum $Z$ as a complex geometric series.
- Rationalize and separate $Z$ into real and imaginary parts to recover $S_1$ and $S_2$.
Step 1 · Evaluate the complex geometric series
Define $Z = \sum_{n=1}^{\infty} q^n e^{inx} = \sum_{n=1}^{\infty} q^n(\cos(nx)+i\sin(nx))$. By linearity, $Z = S_2 + iS_1$. Rewriting:
$$ Z = \sum_{n=1}^{\infty} (qe^{ix})^n. $$
This is geometric with first term and ratio $qe^{ix}$; since $|qe^{ix}|=|q|<1$, it converges to:
$$ Z = \frac{qe^{ix}}{1-qe^{ix}}. $$
Step 2 · Simplify the complex expression
Writing $e^{ix}=\cos x+i\sin x$ and multiplying numerator and denominator by the conjugate of the denominator, the denominator becomes:
$$ (1-q\cos x)^2+(q\sin x)^2 = 1-2q\cos x+q^2. $$
The numerator simplifies to:
$$ (q\cos x+iq\sin x)\big[(1-q\cos x)+iq\sin x\big] = q\cos x-q^2 + iq\sin x. $$
Therefore:
$$ Z = \frac{q\cos x-q^2+iq\sin x}{1-2q\cos x+q^2}. $$
Step 3 · Extract real and imaginary parts
Reading off the real and imaginary components of $Z$:
$$ \operatorname{Re}(Z) = \frac{q\cos x-q^2}{1-2q\cos x+q^2} = S_2, \qquad \operatorname{Im}(Z) = \frac{q\sin x}{1-2q\cos x+q^2} = S_1. $$
Summary Table
| Series | Closed form |
|---|---|
| $S_1 = \sum q^n\sin(nx)$ | $\dfrac{q\sin x}{1-2q\cos x+q^2}$ |
| $S_2 = \sum q^n\cos(nx)$ | $\dfrac{q\cos x-q^2}{1-2q\cos x+q^2}$ |
Additional Notes
- The shared denominator $1-2q\cos x+q^2$ factors as $(1-qe^{ix})(1-qe^{-ix}) = |1-qe^{ix}|^2$, which is strictly positive for $|q|<1$, so both formulas are always well-defined and free of singularities.
- Setting $x=0$ gives $S_2 = \sum q^n = \frac{q-q^2}{1-2q+q^2} = \frac{q(1-q)}{(1-q)^2} = \frac{q}{1-q}$ and $S_1=0$, matching the ordinary geometric series result from Problem 1.
- This complex-exponential technique generalizes readily: any series of the form $\sum q^n\sin(nx)$ or $\sum q^n\cos(nx)$ (or with polynomial factors of $n$ inserted) can be tackled the same way by first summing the corresponding complex geometric or power series.
Result
Series $1^\circ$:
Series $2^\circ$: