$(c\_k)$ is real number sequence. It is known that $\lim\_{y\to 1^-}\sum\_{k=1}^{\infty} c\_k y^k $ exists. Moreover, $\lim\_{k\to \infty} kc\_k=0$ and $\sum\_{k=1}^{\infty}c\_ky^k$ exists for all $y \in [0,1)$.Suppose $\lim\_{n\to\infty}\frac{1}{n}\sum \_{k=1}^n kc\_k=0$ 


ٍٍٍٍٍٍٍٍٍٍٍٍٍSuppose that for every $y \in (0,1)$ we take G(y), which is a natural number and depends on y, such that $$ G(y) \leq \frac{1}{1-y} \leq 1+G(y) .$$ 


I want to prove that $\lim\_{y\to 1^-}\sum\_{k=1}^{G(y)} c\_k(1-y^k) =0$. Then finally I want to prove that $\lim\_{y\to 1^-}\sum\_{k=1}^{G(y)} c\_k=\lim\_{y\to1^-} \sum\_{k=1}^{\infty}c\_ky^k$. 


I tried the following: 


For $0\leq y < 1$, the sequence $(y\_n)$ is monotonically decreasing and bounded, and therefore the Abel test can be used . Then, $\sum\_{n=1}^{\infty}c\_{2n}y^{2n}$ and $\sum\_{n=1}^{\infty}c\_{2n-1}y^{2n-1}$ are convergent due to $\sum\_{n=1}^{\infty}c\_{n}y^{n}=\sum\_{n=1}^{\infty}c\_{2n}y^{2n}+\sum\_{n=1}^{\infty}c\_{2n-1}y^{2n-1}$. But then I feel something is wrong.
I am confused. I want to do it using elementary analysis. Please help!



Convergence of infinite series
Note that the following asymptotic expansion holds for $x\simeq0$
$$
(x+1)^k-1=kx+\mathcal{O}(x^2).
$$ 
Using the above with $x=1-y$ and recalling that $G(y)\simeq (1-y)^{-1}$, we get for $y\simeq 1$
$$
\sum\_{k=1}^{G(y)}c\_k(1-y^k)=(1-y)\sum\_{k=1}^{G(y)}k c\_k+\mathcal{O}\left((1-y)^2 \sum\_{k=1}^{G(y)} c\_k\right)=(1-y)\sum\_{k=1}^{G(y)}k c\_k+\mathcal{O}\left(1-y\right).
$$
Noticing that as $y\rightarrow 1^{-}$ the first term on the right hand side behaves like
$$
\lim\_{n\rightarrow\infty}\frac{1}n \sum\_{k=1}^n k c\_k=\lim\_{n\rightarrow\infty} kc\_k=0
$$
we finally get
$$
\lim\_{y\rightarrow 1^{-}}\sum\_{k=1}^{G(y)}c\_k(1-y^k)=0.
$$
From it follows that
$$
\lim\_{y\rightarrow 1^{-}}\sum\_{k=1}^{G(y)}c\_k=\lim\_{y\rightarrow 1^{-}}\sum\_{k=1}^{G(y)}c\_k y^k,
$$
due to the assumption that $\lim\_{y\rightarrow 1^{-}}\sum\_{k=1}^\infty c\_k y^k$ exists.


