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)

I am not sure that I arrive to the same result $$g(k\pi)=\sum_{n=1}^{\infty}\frac{1}{10^{n}}=\frac{1}{9}$$ 
Using complex representation of $\sin(x)$, what I got is that $$g\left ( x \right )= \sum_{n=1}^{\infty}\frac{\sin \left ( nx \right )}{10^{n}\sin \left ( x \right )}=-\frac{10 e^{i x}}{\left(-10+e^{i x}\right) \left(-1+10 e^{i x}\right)}$$ So, $g\big(2k\pi\big)=\frac{10}{81}$, $g\big((2k+1)\pi\big)=\frac{10}{121}$.
