Consider a sequence of power series, indexed by $k$, as $f_k(\beta) = \sum_{j= 0}^{[k/2]} (-1)^j \binom{k}{2j} \beta^{2k-2j}$, where $[k/2]$ denotes the largest integer that is smaller than $k$.
I'm interested in the asymptotic properties of $f_k$. Specifically, I conjecture that $\lim_{k\to \infty} f_k(\beta) = 0$ for all $\beta\in (-1, 1)$. Can anyone help me prove whether it is correct?
This conjecture seems false. With $i=\sqrt{-1}$, note that \begin{align*} (\beta+i)^k = \sum_{n=0}^{k} \binom{k}{n} i^n \beta^{k-n} = \sum_{j=0}^{\lfloor k/2\rfloor} \binom{k}{2j} (-1)^j \beta^{k-2j} + i \sum_{j=0}^{\lfloor(k-1)/2\rfloor} \binom{k}{2j+1} (-1)^j \beta^{k-2j-1}. \end{align*} In particular, $$ f_k(\beta) = \beta^k \mathop{\rm Re}\big( (\beta+i)^k \big). $$ Then $$ |f_k(\beta)| \le |\beta^k(\beta+i)^k| = \big(|\beta|\sqrt{\beta^2+1}\big)^k, $$ which shows that $\lim_{k\to\infty} f_k(\beta) = 0$ when $|\beta|\sqrt{\beta^2+1} < 1$, or in other words when $|\beta| < \sqrt{(\sqrt5-1)/2}$. However, for larger values of $|\beta|$, this maximum value grows as $k$ increases; and indeed $\beta^k \mathop{\rm Re}\big( (\beta+i)^k \big)$ will be close to $2^{k/2}$ when $k$ is divisible by $8$ and $\beta$ is near $1$, for example.