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.
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.