Series with limit

Question

The problem is the following. Let $(a_{n})_{n\geq0}$ such that $\lim_{n\to\infty}a_{n}=\alpha\in\mathbb{C}$ and let $(b_{n})_{n\geq0}$ a succession of positive real numbers. We know that the series $$ \sum_{n=0}^{\infty}b_{n}z^{n} $$ converges for every $z\in\mathbb{C}$. Prove that, considering now $z\in\mathbb{R}$, $$ \lim_{z\to+\infty}\frac{\sum_{n=0}^{\infty}a_{n}b_{n}z^{n}}{\sum_{n=0}^{\infty}b_{n}z^{n}}=\alpha. $$ If one could switch the limit and the series, then it would be done, but it doesn't seem so easy. How can be proved the limit otherwise?

Answer 1

Suppose $\alpha = 0.$ Let $\epsilon>0.$ Then there exists $N$ such that $|a_n|<\epsilon$ for $n>N.$ The quotient of interest can be written

$$\frac{\sum_{n=0}^N a_nb_nz^n + \sum_{n=N+1}^\infty a_nb_nz^n}{\sum_{n=0}^N b_nz^n + \sum_{n=N+1}^\infty b_nz^n}.$$

For $z>0,$ the absolute value of this is bounded above by

$$\frac{\sum_{n=0}^N |a_n|b_nz^n + \epsilon\sum_{n=N+1}^\infty b_nz^n}{\sum_{n=0}^N b_nz^n + \sum_{n=N+1}^\infty b_nz^n}.$$

Now divide top and bottom by $\sum_{n=N+1}^\infty b_nz^n$ and take the $\limsup$ as $z\to \infty.$ We see this $\limsup$ is no more than

$$\frac{0+\epsilon}{0+1} = \epsilon.$$

This takes care of the $\alpha=0$ case, and if you think about it, gives all other cases as well.