Proving a function is continuous and periodic Suppose we are given a function 
$$g\left ( x \right )= \sum_{n=1}^{\infty}\frac{\sin \left ( nx \right )}{10^{n}\sin \left ( x \right )},x\neq k\pi , k\in\mathbb{Z}$$
and 
$$g\left ( k\pi \right )=\lim _{x\rightarrow k\pi}g\left ( x \right )$$
I found that $\lim _{x\rightarrow k\pi}g\left ( x \right )= \sum_{n=1}^{\infty}\frac{1}{10^{n}}$ for both odd and even $k$. However, I am still unsure how to proceed from here. What does continuity even mean for functions like this? If I wanted to take the derivative of this function, surely I must prove first that it converges and then I would apply differentiation term by term. 
Proving periodicity for one term would be easy enough if it weren't for the $10^{n}$ term in the denominator. How to deal with it? 
EDIT: How to find the Fourier series of this function( by first proving that Dirichlet's conditions are met)

It is a basic precalculus fact that for arbitrary complex $z$ with $|z|<1$ one has
$$\sum_{k=0}^\infty z^k={1\over 1-z}\ .$$
Furthermore, by Euler's equation, for arbitrary real $x$ we may write $\sin x={\rm Im}\>e^{ix}$ . Putting these two together we obtain
$$\eqalign{\sum_{n=1}^\infty{\sin (nx)\over 10^n}&={\rm Im}\sum_{n=1}^\infty\left({e^{ix}\over10}\right)^n={\rm Im}{e^{ix}/10\over 1-e^{ix}/10}\cr &={\rm Im}{e^{ix}(10-e^{-ix})\over (10-e^{ix})(10-e^{-ix})}={10\sin x\over101-20\cos x}\ .\cr}$$
It follows that
$$g(x)={10\over 101-20\cos x}\qquad(x\in{\mathbb R}\setminus \pi{\mathbb Z})\ .\tag{1}$$
It follows that the apparent singularities at integer $x$ are removable, so that $g$ has a natural $C^\infty$ extension to all of ${\mathbb R}$, and this extension is $2\pi$-periodic. Note that the initially given expression for $g$ is not its Fourier series. In order to obtain the latter one would have to compute the integrals $$a_k:={1\over\pi}\int_0^{2\pi}g(x)\>\cos(kx)\>dx\ ,$$ using the expression $(1)$ for $g$. The resulting expressions for the $a_k$ will not be simple.
