Absolute and Uniform Convergence of a Series of Functions

Question

Consider the series

$$F(x)=\sum_{n=1}^{\infty}\frac{\cos(nx)}{n}$$

with $0<a\le x\le b<2\pi$

Show that over $[a,b]$, $F$ is uniformly, but not absolutely, convergent.

Using Dirichlet's test, I assume we take b_n=1/n. Then all we need to show is that the partial sums of cos(nx) are bounded. I'm not exactly sure how to prove that. Alternatively, we could use the other reduction, and solve the problem quite easily.
Do either methods help with absolute convergence? — ant11, Feb 23 '14 at 22:33
See the last link I gave. The absolute value of the expression given there is bounded by $1/|\sin(x/2)|$. This is bounded above for $x$ in your given range. — David Mitra, Feb 23 '14 at 22:36
For absolute convergence, you probably need to break into two cases: when x is a rational multiple of $2\pi$, versus when $x/(2\pi)$ is irrational. The first case can be easily handled; for the second, use Weyl equidistribution theorem. — John Jiang, Feb 23 '14 at 22:37

score 1 · Answer 1 · answered Feb 23 '14 at 22:53

For uniform convergence use Dirichlet's criterio, as indicated by David's commnet.

For non absolute convergence:

$$ \sum\frac{|\cos(n\,x)|}{n}\ge\sum\frac{|\cos(n\,x)|^2}{n}=\sum\frac{1+\cos(2\,n\,x)}{2\,n}=\sum\frac{1}{2\,n}+\sum\frac{\cos(2\,n\,x)}{2\,n}. $$ The first series is divergent, while the second is convergent. In fact, this shows that $$ \sum_{n=1}^N\frac{|\cos(n\,x)|}{n}\ge\frac{\log N}{2}-C $$ for some constant $C>0$.

score 1 · Answer 2 · answered Feb 23 '14 at 23:13

It has already been observed that the series does not converge absolutely.

Next, we show that it converges uniformly in $[a,b]$.

First we find a uniform bound for the sequence: $$ s_n(x)=\cos x+\cos 2x+\cdots+\cos nx=\mathrm{Re}\,\left(\mathrm{e}^{ix}+\mathrm{e}^{i2x}+\cdots+\mathrm{e}^{inx}\right)=\mathrm{Re}\,\frac{\mathrm{e}^{i(n+1)x}-\mathrm{e}^{ix}}{\mathrm{e}^{ix}-1}, $$ and $$ \frac{\mathrm{e}^{i(n+1)x}-\mathrm{e}^{ix}}{\mathrm{e}^{ix}-1}=\frac{\mathrm{e}^{i(n+1/2)x}-\mathrm{e}^{ix/2}}{\mathrm{e}^{ix/2}-\mathrm{e}^{-ix/2}}=\frac{\mathrm{e}^{i(n+1/2)x}-\mathrm{e}^{ix/2}}{2i\sin(x/2)} $$ and thus $$ \lvert s_n(x)\rvert\le \frac{1}{\lvert\sin(x/2)\rvert}\le\max_{x\in [a,b]}\frac{1}{\lvert\sin(x/2)\rvert}=M. $$ Note that $M<\infty$, since $0<a\le b<2\pi$.

Next we use the Abel's summation method \begin{align} \sum_{k=1}^n\frac{\cos nx}{n}&=\sum_{k=1}^n\frac{s_n(x)-s_{n-1}(x)}{n}= \sum_{k=1}^n\frac{s_n(x)}{n}-\sum_{k=1}^n\frac{s_{n-1}(x)}{n} \\&= \sum_{k=1}^n\frac{s_n(x)}{n}-\sum_{k=1}^{n-1}\frac{s_{n}(x)}{n+1} =\frac{s_n(x)}{n}+\sum_{k=1}^{n-1}s_n(x)\left(\frac{1}{n}-\frac{1}{n+1}\right) \\ &=\sum_{k=1}^{n-1}\frac{s_n(x)}{n(n+1)}+\frac{s_n(x)}{n}, \end{align} which shows that our series converges uniformly as $$ \left\vert\frac{s_n(x)}{n}\right\rvert\le \frac{M}{n}\to 0, $$ and the series $$ \sum_{k=1}^{n-1}\frac{s_n(x)}{n(n+1)}, $$ converges uniformly (and absolutely) due to comparison test since $$ \left\vert\frac{s_n(x)}{n(n+1)}\right\rvert\le \frac{M}{n(n+1)} \quad\text{and}\quad \sum_{n=1}^\infty \frac{M}{n(n+1)}<\infty. $$

score 0 · Answer 3 · answered Feb 23 '14 at 22:46

Use the trigonometric identity

$$2\cos A\sin B=\sin(A+B)-\sin(A-B)$$

with, as indicated by DavidMitra, $A=nx$, $B=\frac12 x$.

Then

$$2\sin(\tfrac12 x)\sum_{n=1}^N\frac{\cos(nx)}{n}=-\sin(\tfrac12 x)+\sum_{n=1}^{N-1}\frac{\sin((n+\tfrac12) x)}{n(n+1)}+\frac{\sin((N+\tfrac12) x)}{N}$$

where now uniform convergence of the middle sum follows from the classical majorant $\sum\frac1{n(n+1)}$.

Absolute and Uniform Convergence of a Series of Functions

3 Answers3