Regarding expressing Lambert series in terms of Dirichlet Convolution I am studying Lambert Series . 
It's definition says a series of the form $\sum_{n=1}^\infty  \frac { f(n) x^n } { 1 -  x^n }  $ = 
 $\sum_{n=1}^\infty  F(n) x^n $ , where $F(n) =  \sum_{d|n}  f(d) $ . 

I can think about LHS of defination equal to RHS if $|x|<1$ and then expanding $\frac{1} {1-x^n } $ = 1 + $x^n $ + ....    . 
  But can someone prove it by more elegence using properties of Dirichlet Convolution or something else rather than brute force. 

Edit $1$ - In calculating LHS I wrote $d | n$ to $n= m × d$ and then as $n$ approaches = $\infty $ both my and d must approach $\infty $  and then collecting powers of $x$ but I think not a good way to prove it. Can someone please give another and nice proof. 
 A: Here is one possible approach. Given two sequences of numbers $\,a_1,a_2,\dots\,$
and $\,b_1,b_2,\dots,\,$ then their Dirichlet convolution is defined as
$$ c_n = (a*b)_n := \sum_{d|n} a_d\,b_{n/d}
= \sum_{ij=n} a_ib_j. \tag{1}$$
Consider the ordinary generating function
of the $\,a, b\,$ sequences defined by
$$ f(x):=\sum_i a_i\,x^i, \;\;
g(x):=\sum_j b_j\,x^j \tag{2} $$
where all summations are over positive integers only.
The o.g.f. of their Dirichlet convolution
$\,c\,$ is given by
$$ h(x)\! :=\! \sum_{i,j} a_ib_jx^{ij} \!=\! 
\sum_j f(x^j)b_j \!=\! 
\sum_i a_ig(x^i). \tag{3} $$
This is equivalent to summing a matrix first by
rows and then columns giving the same result
as summing by columns first.
A special case is the constant sequence $\,b_n = 1\,$ whose o.g.f is
$$ g(x) = \sum_j x^j = \frac{x}{1-x} \tag{4} $$ and where the convolution is
$\, c_n = \sum_{i|n} a_i.\,$ 
Thus, using equation $(3)$ we get immediately that
$$ h(x) \!=\! \sum_{i,j} a_ix^{ij} \!=\!
 \sum_j f(x^j) \!=\!
 \sum_i a_i\,\frac{x^i}{1\!-\!x^i}. \tag{5} $$
which is a Lambert series.
