# Show that the function $\cos(x)\sum_{k=0}^\infty \Bigl(\prod_{l=1}^{k}\frac{2l}{2l+1}\Bigr)\sin^{2k+1}(x)$ is continuous

Can you prove or disprove the following?

The periodic function $$f:[0,2\pi]\to \mathbb{R}$$ \begin{align*} f(x):=\cos(x)\sum_{k=0}^\infty \bigg(\prod_{l=1}^{k}\frac{2l}{2l+1}\bigg)\sin^{2k+1}(x) \end{align*} is continuous and attains therefore its maximum on the interval $$[0,2\pi]$$. Just to make notation clear: The empty product $$\prod_{l=1}^0$$ is set to 1.

For $$x\ne\frac{\pi}{2}$$, $$f$$ is continuous because the sum converges. This follows since the geometric series $$\sum_{k=0}^\infty \sin^k(x)$$ converges. We have to show continuity for $$x=\frac{\pi}{2}$$.

EDIT 2:

We claim that it is possible to show that \begin{align*} \cos(x)\sum_{k=0}^\infty \bigg(\prod_{l=1}^{k}\frac{2l}{2l+1}\bigg) \sin^{2k+1}(x)\leq \cos(x)\sum_{k=0}^\infty \frac{\sin^{2k+1}(x)}{\sqrt{2k+1}}=:g(x). \end{align*} Hence, it is left to prove that $$g(x)$$ is contiuous in $$\frac{\pi}{2}$$.

It is known that the function \begin{align*} \cos(x)\sum_{k=0}^\infty \sin^k(x)=\begin{cases} \frac{\cos(x)}{1-\sin(x)} & \text{for }\quad x\ne \frac{\pi}{2}\\ 0 & \text{for }\quad x= \frac{\pi}{2} \end{cases} \end{align*} is not continous.

Notice, in EDIT 1, there was a mistake. It should be but the square root as it is now, not the k-root.

• EDIT: It is sufficient to show the function is continuous in $\frac{\pi}{2}$ – user3154270 Jun 9 at 21:50

Maybe the Cauchy product helps. Define $$h(x)=\sum_{k=0}^\infty a_k = \sum_{k=0}^\infty \bigg(\prod_{l=1}^{k}\frac{2l}{2l+1}\bigg)\sin^{2k+1}(x)$$ and $$\cos(x)=\sum_{k=0}^\infty b_k=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}$$ Then, $$\cos(x)h(x) = \sum_{k=0}^\infty a_k \sum_{k=0}^\infty b_k = \sum_{k=0}^\infty c_k$$ where $$c_k=\sum_{i=0}^k a_i b_{k-i}$$.
We calculate $$c_k$$ in this case $$c_k=\sin^{2k+1}(x)\sum_{i=0}^k (-1)^i\frac{x^{2i}}{\sin^{2i}(x)}\frac{1}{(2i)!}\bigg(\prod_{l=1}^{k-i}\frac{2l}{2l+1}\bigg).$$ The hardest problem is now to show the convergence of $$\sum_{k=0}^\infty c_k$$.
Since is it known that $$\cos(x)h(x)$$ is continous for $$x<\frac{\pi}{2}$$, it is sufficient to show that the series $$\sum_{k=0}^\infty c_k$$ converges for $$x=\frac{\pi}{2}$$.
• Maybe $\lim_{x\to o}\frac{sin(x)}{x}=1$ helps – user3154270 Jun 9 at 21:58