$\int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx$ by changing integral and summation.

Question

Let $f(x)=\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}$. Prove that $f$ converges uniformly on $[0,2\pi]$ and evaluate $\int _0^{2\pi}f(x)^2dx$.

Since $\left|\frac{\sin nx}{2^n}\right|\le\left|\frac{1}{2^n}\right|$ and $\sum \limits _{n=1}^\infty \frac{1}{2^n}$ converges, $f(x)$ converges uniformly on $[0,2\pi]$.

After reading some related post, Changing integral and summation, and Calculate $\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx$. I need to prove $$\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 \text{converges uniformly on }[0,2\pi].$$

Write $\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2=\sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \frac{\sin nx}{2^n}\frac{\sin mx}{2^m}=\sum_\limits{k=1}^\infty c_k$, where $c_k=\sum_\limits{j=1}^k\frac{\sin jx\sin (k-j)x}{2^k}$.

So $$|c_k|\le\left|\frac{k}{2^k} \right|$$ Is $\sum_\limits{n=1}^\infty \frac{k}{2^k}$ converges?

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I have another question. In the linked post, they write $$\begin{align} \int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx & = \int _0^{2\pi} \sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \frac{\sin nx}{2^n}\frac{\sin mx}{2^m} dx \\&=\sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \int _0^{2\pi} \frac{\sin nx}{2^n}\frac{\sin mx}{2^m}dx . \end{align}$$

But I think it is a Cauchy product and so $$\begin{align} \int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx &=\int _0^{2\pi}\sum_\limits{k=1}^\infty \sum_\limits{j=1}^k\frac{\sin jx\sin (k-j)x}{2^k}dx . \end{align}$$

What is the difference between these two integrals?

Answer 1

(1). A useful lemma (which has a short elementary proof) is that if $g_n\to g$ uniformly on a domain $D\subseteq \Bbb R$ and if each $g_n$ is continuous on $D$ then $g$ is continuous on $D.$ From this, if, also, $D$ is a closed bounded interval, then $g_n^2\to g^2$ uniformly on $D,$ as follows:

Let $M=\sup_{x\in D}|g (x)|.$ Then $M\ne \infty$ because $D$ is a closed bounded interval and $g$ is continuous on $D.$

Let $n_0\in\Bbb N$ such that $\forall n\ge n_0 \,(\sup_{x\in D}|g_n(x)-g(x)|\le 1).$ Then $\forall n\ge n_0 \,(\sup_{x\in D} |g_n(x)|\le 1+M).$

So if $n\ge n_0$ and $x\in D$ then $$|g_n(x)^2-g(x)^2|=|g_n(x)-g(x)|\cdot |g_n(x)+g(x)|\le$$ $$\le |g_n(x)-g(x)|\cdot (|g_n(x)|+|g(x)|) \le$$ $$\le |g_n(x)-g(x)|\cdot (2M+1).$$ So for all but finitely many $n$ we have $$\sup_{x\in D} |g_n(x)^2-g(x)^2|\le (2M+1)\cdot \sup_{x\in D}|g_n(x)-g(x)|.$$

(2). In your Q, with $D=[0,2\pi],$ let $g_n(x)=\sum_{j=1}^{n}(\sin jx)/2^j.$ Then $g_n(x)$ converges to a value $g(x)$ for each $x$ by term-by-term comparison with the absolutely convergent series $\sum_{j=1}^{\infty}1/2^j.$ And $g_n$ converges uniformly because $\sup_{x\in D}|g_n(x)-g(x)|=$ $=\sup_{x\in D}|\sum_{j=n+1}^{\infty}(\sin jx)/2^j|\le$ $ \sum_{j=n+1}^{\infty}1/2^j=1/2^n.$ And each $g_n$ is continuous.

Therefore by (1), $g_n^2\to g^2$ uniformly.

Remark: In (1), to obtain uniform convergence of $g_n^2$ it is usually necessary that $D$ is closed and bounded. Examples: (i). $D=\Bbb R$ and $g_n(x)=x+1/n$ and $g(x)=x.$ (ii). $D=(0,1]$ and $g_n(x)=1/x +1/n$ and $g(x)=1/x.$ In both (i) and (ii), $g_n\to g$ uniformly but $\sup_{x\in D}|g_n(x)^2-g(x)^2|=\infty$ for every $n\in \Bbb Z^+.$

Answer 2

Yes, the series converges. You showed that $|c_k|\le k/2^k$. Note that for $|q|>1$ we have $$\sum_{n\ge1}\frac{1}{q^n}=\frac{1}{q-1},$$ from the geometric series (which converges uniformly). Then, $$\sum_{n\ge1}\frac{n}{q^n}=-\sum_{n\ge1}q\frac{d}{dq}q^{-n}=-q\frac{d}{dq}\left(\sum_{n\ge1}\frac1{q^n}\right)=-q\frac{d}{dq}\left(\frac1{q-1}\right)=\frac{q}{(q-1)^2}.$$ Remember, this is for any $q$ with $|q|>1$, so we can choose $q=2$ and get $$\sum_{n\ge1}|c_n|\le\sum_{n\ge1}\frac{n}{2^n}=2.$$