Power series with alternating signs and its zeros. This may be a strange question, but I've not found anything about this.
Well, anyone can observe that both
$$
\cos(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n)!}z^{2n}
$$
and
$$
\sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}z^{2n+1}
$$
are alternating power series and that they all the roots on the real axis.
My question:


*

*Does the alternating sign (of real coefficients), somehow, implies that all the zeros are real?

*If not, are there alternating power series with complex zeros, I mean counterexamples?


Thanks.
 A: An analytic function may have coefficients with alternating signs and still violate Newton's inequalities, enforcing a complex root. For instance,
$$ f(x) = x^2+\sum_{k\geq 0}\frac{(-1)^k}{k!}x^k = e^{-x}+x^2 $$
has complex roots at $2\cdot W\left(\pm\frac{i}{2}\right)\approx 0.325199\, \pm 0.785257\, i.$
A: 1. No. 2. Yes. Consider the function $f$ defined on $\mathbb{C}$ by the power series 
$$f(z) = \sum_{n=0}^\infty (-1)^n a_n z^n$$
where $a_0=a_2=1$, $a_1=0$, and $a_n=0$ for all $n\geq 2$ (so that, indeed, $a_n \geq 0$ for all $n\in\mathbb{N}$). We do have that this series is alternating... even though it is a bit trivial.
Now, what are the zeroes of $f(z) = 1+z^2$?
A: Consider $$f(z) = \frac{1-z+z^2}{4+z}= \frac14 - \frac{5}{16}z + \sum_{n>1} 
(-1)^n\frac{21}{2^{2n+2}}z^n$$
The terms in its expansion about $z=0$ are of alternating signs.  Yet it has two complex zeros, at 
$$
\frac{1\pm\sqrt{3}i}{2}
$$
A simpler example is 
$$g(z) = \frac{1-z+z^2}{1+z}= 1 - 2z + \sum_{n>1} 
(-1)^n3z^n$$
A: A more general method:  Start with an alternating series $\sum_{k} (-1)^k a_k x^k$ (for simplicity of exposition, suppose $a_1 > 0$) for which all the zeroes lie on the real line.  Now, add a multiple of $x$, deforming the locations of any surviving zeroes away from the real line (because a zero at $z$ now has a multiple of $z$ as its value).  Or, start over and subtract a multiple of $x^2$, deforming the locations of any surviving zeroes away from the real line (because a zero at $z$ now has a multiple of $z^2$ as its value).  Having zeroes on the real line is a delicate thing, which we can break easily.
Following the pattern here:  increasing the first positive coefficient in the power series for sine, $x + \sin x$ only has one zero on the real line, at $0$.  It has more, for instance, at $4.21239... + 2.25073... \mathrm{i}$.
