Prove that $0<\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}$ for $x>0$ So far, I've tried using induction:
Let P(n): $0<\sum_{i=2}^n\ (-1)^i \frac{x^{2i}}{2i!}$
Let $n=2$. Thus, P(2): $\frac{x^4}{4!}>0$ for $x>0$. Then, let P(k) be true (Induction Hypothesis). Now, prove the validity of P for n=k+1:
$P(k+1) : \sum_{i=2}^{k+1}\ (-1)^i\frac{x^{2i}}{2i!} = \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{k+1}\frac{x^{2(k+1)}}{2(k+1)!} = \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} - (-1)^{k}\frac{x^{2(k+1)}}{2(k+1)!}$
Let $k=2m-1$:
$\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} - (-1)^{k}\frac{x^{2(k+1)}}{2(k+1)!}=\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{2m}\frac{x^{4m}}{4m!}$ which is $>0$ since $\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!}>0$ and $\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{2m}\frac{x^{4m}}{4m!} > \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!}$ for $x>0$ according to the Induction Hypothesis.
I'm lost for the case where $k=2m$, I'd like some help there, please.
 A: This isn't true for each $n$. For example, when $x=10$ and $n=5$
$$\sum_{i=2}^5(-1)^{i}\frac{10^{2i}}{(2i)!}=-\frac{707500}{567}<0 $$
However, for a fixed $x$, it will be true for sufficiently large $n$. This is because $\sum_{i=2}^n(-1)^{i}\frac{x^{2i}}{(2i)!}$ converges to $\cos x-1+x^2/2$ which is positive for $x\neq 0$, and so there's an $N\in\mathbb{N}$ such that for $n\geq N$, we can say that $\sum_{i=2}^n(-1)^{i}\frac{x^{2i}}{(2i)!}>0$. Note the $N$ depends on the value of $x$, and I don't think there's one independent of $x$.
A: $$S_n=\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}=\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}-\sum_{i=0}^2\ (-1)^n \frac{x^{2n}}{2n!}$$
$$S_n=\frac{x^2-1}2+\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}$$ If you are aware of the complete and incomplete gamma function
$$\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}=\frac{e^{-x^2}\,\, \Gamma \left(n+1,-x^2\right)}{2 \Gamma (n+1)}$$
$$S_n=\frac{1}{2} \left(e^{-x^2}\frac{ \Gamma \left(n+1,-x^2\right)}{\Gamma
   (n+1)}+x^2-1\right)$$ which, for a given $n$ or a given $x$ can cancel.
As @bjorn93 did, let us use $x=10$ and $n=5$. $S_5$ will become negative as soon as $x>1.96259$. Checking for $x=2$, we have $-\frac{4}{15}$
