Limit of given expression Let $\sum a_k=s$. I want to show that $$\lim\limits_{x\to 1^-}(1-x)\sum\limits_{k=1}^{\infty}\frac{ka_kx^k}{1-x^k}=s$$ where $x\in(0,1)$. Thanks for your helps.
 A: The initial version of this proof was mistaken. All credit for this one should go to @grizzly , who suggested this approach in a comment below.
Note that $$\lim \limits _{x \to 1^-} (1-x) \sum \limits _{k=1} ^\infty \frac {k a_k x^k} {1-x^k} = \lim \limits _{x \to 1^-} \sum \limits _{k=1} ^\infty \frac {(1-x) k a_k x^k} {(1-x) (1 + x^2 + x^3 + \cdots + x^{k-1} )} = \\ \lim \limits _{x \to 1^-} x \sum \limits _{k=1} ^\infty \frac {k a_k x^{k-1}} {1 + x^2 + x^3 + \cdots + x^{k-1}} = \lim \limits _{x \to 1^-} \sum \limits _{k=1} ^\infty \frac {k a_k x^{k-1}} {1 + x^2 + x^3 + \cdots + x^{k-1}} .$$
It would be tempting to interchange the limit and the sum and continue with $\sum \limits _{k=1} ^\infty \lim \limits _{x \to 1^-} \frac {k a_k x^{k-1}} {1 + x^2 + x^3 + \cdots + x^{k-1}}$, after which things would be really easy. Unfortunately, it is not obvious that the interchange is possible, and in fact there are enough examples in mathematics where it isn't. One sufficient condition guaranteeing that you may do it is for the series to converge uniformly - which is exactly what we shall try to prove.
Let $\alpha _k (x) = a_k :(0,1) \to \Bbb R$. Note that $\sum \limits _{k=1} ^\infty \alpha _k (x)$ converges uniformly (to $s$), being a series of constant functions.
Let $f_k (x) = \frac {k x^{k-1}} {1 + x^2 + x^3 + \cdots + x^{k-1}} : (0,1) \to 
\Bbb R$. Note that since $0<x<1$, you have $1 + x + x^2 + \cdots + x^{k-1} \ge k x^{k-1}$, therefore $0 \le f_k \le 1$, so that the functions $f_k$ are uniformly bounded.
Furthermore, it is easy to check by hand that $f_{k+1} \le f_k$ (this is equivalent to $x (1+x+x^2+ \cdots + x^{k-1}) \le k$ which is obviously true).
One may now apply the variant of Abel's uniform convergence test found in exercise 9.3.26 of "Elementary Real Analysis" by Brian S. Thomson, Judith B. Bruckner, Andrew M. Bruckner (2nd edition) to conclude that the series given in the problem converges uniformly and you are allowed, then, to interchange summation and limit. Doing that amounts to replacing $x$ by $1$, giving $\sum \limits _{k=1} ^\infty \frac {k a_k} {k} = s$.
