Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{\cos nx}{e^{nx}}. $$
Explanation of Method
Recognizing $\cos(nx)=\mathrm{Re}(e^{inx})$ turns the series into the real part of a complex geometric series with ratio $z=e^{-x+ix}$, whose magnitude $|z|=e^{-x}$ controls convergence. This gives the threshold $x>0$ directly, matching the direct comparison $|\cos(nx)/e^{nx}|\le e^{-nx}$.
- Express the series as the real part of a complex geometric series.
- Determine the convergence threshold from $|z|=e^{-x}<1$.
- Confirm absolute convergence for $x>0$ and divergence for $x\le0$ directly.
Step 1 · Complex geometric series
$\displaystyle\sum_{n=1}^{\infty}\frac{\cos nx}{e^{nx}} = \mathrm{Re}\sum_{n=1}^{\infty}\left(e^{ix}e^{-x}\right)^n = \mathrm{Re}\sum_{n=1}^{\infty}z^n$, with $z=e^{-x+ix}$, $|z|=e^{-x}$.
Step 2 · Threshold
The geometric series $\sum z^n$ converges iff $|z|=e^{-x}<1$, i.e. $x>0$.
Step 3 · Confirm directly
For $x>0$: $\left|\dfrac{\cos nx}{e^{nx}}\right|\le e^{-nx}$, and $\sum e^{-nx}$ converges (geometric, ratio $e^{-x}<1$) — absolute convergence. For $x\le0$: $e^{-nx}\ge1$ for every $n$, so $|\cos(nx)/e^{nx}|$ does not tend to $0$ in general (and at $x=0$ every term equals $1$), so the series diverges by the $n$-th term test.
Summary Table
| Region | Behavior |
|---|---|
| $x>0$ | Absolutely convergent |
| $x\le0$ | Diverges |
Additional Notes
- This is the exponentially-damped cousin of Problem 133's $\sum\sin(nx)/n^a$: there, the oscillation's boundedness alone (via Dirichlet's test) sufficed for convergence with a merely polynomial weight; here, the exponential weight $e^{-nx}$ is strong enough to secure absolute convergence outright once $x>0$, without any Dirichlet-type argument needed.
- The closed form (real part of the geometric sum $z/(1-z)$) works out to $\dfrac{e^{-x}(\cos x-e^{-x})}{1-2e^{-x}\cos x+e^{-2x}}$ for $x>0$, a rational combination reminiscent of the Poisson-kernel formulas in Fourier analysis.
- Since the bound $|\cos(nx)/e^{nx}|\le e^{-nx}$ is completely independent of the oscillation in the numerator, this series' convergence region depends only on the sign of $x$, not on any interplay between $x$'s trigonometric and exponential roles — a simpler structure than the "coupled" convergence problems like 134 or 135.
Result
Converges absolutely for $x>0$; diverges for $x\le0$: