Showing an infimum of an expression involving alternating series is positive in a given variable as another variable tends to infinity Let $\Bbb D$ be the unit disc in the complex plane. For the last month or so, I have been trying to prove that, if $\lambda \in \Bbb D$, and if $n \geq 4$ is an integer,
$$\inf_{\lambda \in \Bbb D}\Big[\frac{1+|\lambda|^2}{2} + Re(\lambda)\Big[1 -  \frac{(2\pi)^{2}}{(n+2)(n+3)}+\frac{(2\pi)^{4}}{(n+2)(n+3)(n+4)(n+5)}-...\Big]-Im(\lambda)\Big[\frac{(2\pi)}{(n+2)} -  \frac{(2\pi)^{3}}{(n+2)(n+3)(n+4)}+...\Big]\Big]$$
$\textit{decreases}$ to $0$ as $n \to \infty$, but that for any fixed $n \geq 4$, this infimum is always positive. I have been able to prove it along the coordinate axes within the disc, and in quadrants 1,3,4 of the disc. What remains is quadrant 2, which for obvious reasons makes the estimate more difficult.
I know that the first series converges to $1$ as $n \to \infty$, and the second series converges to $0$ as $n \to \infty$. I also have used wolfram alpha extensively to work on this, which has shown that for $n$ as large as I have entered,  the statement holds.
Empirically, all the evidence stacks up in my favor, but I can not rigorously show what I need to show here.
 A: The following partial answer only shows that the infimum above is positive, and does so in a bit of a hack-y manner. An elegant answer would be interesting.

Let $\alpha_n$ and $\beta_n$ be the expressions multiplied respectively by $x$ and $y$ in the question. We first show that $\alpha_n^2 + \beta_n^2 < 1$ for $ n \ge 4$.
Note that $\alpha_n$ and $\beta_n$ are defined by alternating sums of monotonically decreasing terms (since $2\pi \le 7 \le n+3$), ergo, they're alternating series. This lets us use the upper bounds by truncation. For instance, $$\alpha_n < 1 -\frac{(2\pi)^{2}}{(n+2)(n+3)}+\frac{(2\pi)^{4}}{\prod_{j = 0}^3(n+2 + j)} $$ $$ \beta_n < \frac{2\pi}{n+2} - \frac{(2\pi)^3}{\prod_{j = 0}^2(n+2 + j)} + \frac{(2\pi)^5}{\prod_{j = 0}^4(n+2 + j)} $$
This form is messy but possible to attack computationally - it merely requires calculating the roots of some $8^{th}$ degree polynomial. Define $$f(n) := \left(1 -\frac{(2\pi)^{2}}{(n+2)(n+3)}+\frac{(2\pi)^{4}}{\prod_{j = 0}^3(n+2 + j)} \right) ^2 + \left(\frac{2\pi}{n+2} - \frac{(2\pi)^3}{\prod_{j = 0}^2(n+2 + j)} + \frac{(2\pi)^5}{\prod_{j = 0}^4(n+2 + j)} \right)^2,$$ and use your favourite package to determine when $f(n) < 1$. For instance, Wolfram Alpha tells me that $f(n) <1$ for $n \ge 3$, and since the bounds above hold for $n \ge 4$, we have that $\alpha_n^2 + \beta_n^2 < f(n) < 1$ for $n \ge 4$. 

Next, we calculate the infimum. Note that
$$ \frac{1 + x^2 + y^2}{2} + \alpha x - \beta y = \frac{1}{2} + \frac{x^2 + 2\alpha x + y^2 - 2\beta y}{2}$$ $$= \frac{1 + (x + \alpha)^2 + (y - \beta)^2}2 - \frac{\alpha^2 + \beta^2}{2}$$ If $\alpha^2 + \beta^2 < 1 $, then the above is minimised at the point $(-\alpha, \beta) \in \mathbb{D}$, since otherwise at least one of $(x + \alpha)^2$ or $(y -\beta)^2$ is positive. Since we know that $\alpha_n^2 + \beta_n^2 <1$, $$ c_n := \inf_\mathbb{D} \frac{1 + x^2 + y^2}{2} + \alpha_n x - \beta_n y = \frac{1 - (\alpha_n^2 + \beta_n^2)}{2} > 0$$

