Can the series f(x)=sum from n=1 to infinity (sin(2^npi
Can the series $\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{\sin(2^n\pi x)}{2^n}$, $x\in(-\infty,\infty)$, be differentiated term by term?
Explanation of Method
The original series converges uniformly by the Weierstrass M-test (majorant $1/2^n$), but that only guarantees continuity of $f$, not differentiability. Term-by-term differentiation additionally requires the series of derivatives to converge (uniformly). Here the derivative series is $\sum\pi\cos(2^n\pi x)$, whose general term does not tend to $0$ at essentially any $x$ — so term-by-term differentiation is not valid.
- Confirm $f$ converges uniformly via the Weierstrass M-test.
- Compute the formal derivative series $\sum\pi\cos(2^n\pi x)$.
- Show this derivative series diverges (its terms fail to vanish), so the theorem cannot be applied.
Step 1 · Uniform convergence of $f$
Since $|\sin(2^n\pi x)|\le1$: $\left|\dfrac{\sin(2^n\pi x)}{2^n}\right|\le\dfrac1{2^n}$, and $\sum1/2^n=1<\infty$. By the Weierstrass M-test, $f$ converges uniformly and absolutely on $\mathbb R$, so $f$ is continuous everywhere.
Step 2 · The formal term-by-term derivative
$u_n'(x)=\dfrac1{2^n}\cdot\cos(2^n\pi x)\cdot2^n\pi=\pi\cos(2^n\pi x)$, so the candidate derivative series is $\sum_{n=1}^{\infty}\pi\cos(2^n\pi x)$.
Step 3 · This series diverges
At $x=0$: $\cos(2^n\pi\cdot0)=\cos0=1$ for every $n$, so the terms are $\pi$ for all $n$ — they do not tend to $0$, so the series diverges at $x=0$ by the Divergence Test. Since term-by-term differentiation requires the derivative series to converge (uniformly) on the whole interval, and it already fails to converge at this single point, the theorem's hypothesis is violated and term-by-term differentiation is not valid.
Summary Table
| Series | Behavior |
|---|---|
| $\sum\sin(2^n\pi x)/2^n$ | Converges uniformly ($M_n=1/2^n$) |
| $\sum\pi\cos(2^n\pi x)$ (formal derivative) | Diverges (e.g. at $x=0$, terms $\equiv\pi$) |
| Term-by-term differentiation | Not valid |
Additional Notes
- This series is a close relative of the classical Weierstrass function $W(x)=\sum a^n\cos(b^n\pi x)$ (continuous everywhere, differentiable nowhere), obtained here with sine instead of cosine and coefficients $a=1/2$, $b=2$ — the divergence of the derivative series is the fingerprint of that same pathology.
- This is the sharpest possible illustration that uniform convergence of $\sum u_n(x)$ alone says nothing about differentiability of the sum: $f$ is continuous everywhere (Step 1), yet the naive term-by-term derivative fails to exist as a series at every point tested.
- Compare with Problems 216 and 217, where an extra power of $1/n$ in the coefficient (rather than a purely geometric one here) is exactly what determines whether the derivative series converges — Problem 216's $\sum\sin(nx)/n$ also fails to differentiate term by term for the same underlying reason (its derivative series $\sum\cos nx$ diverges), while Problem 217's extra $1/n^2$ decay succeeds.
Result
No — the series of derivatives diverges: