<s>[INST] Being careful with terms of infinite sums $ cos(x): = \sum_{k=0}^\infty \frac{(-1)^nx^{2n}}{2n!}$
$=1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}...$
I would like to show that for $x \in [0,2]$
$cos(x) \leq 1- \frac{x^2}{2!}+\frac{x^4}{4!}$
This means the stuff from $-\frac{x^6}{6!}$ onwards must sum to something negative.
So I reckon we can group that stuff into pairs : $(-\frac{x^6}{6!}+\frac{x^8}{8!})+ (-\frac{x^{10}}{10!}+\frac{x^{12}}{12!})...$ and show each bracket is negative.
Indeed : $(-\frac{x^{2n}}{(2n)!}+\frac{x^{2n+2}}{(2n+2)!})= \frac{x^{2n}}{(2n)!}[-1+\frac{x^{2}}{(2n+1)(2n+2)}]$ and the stuff in the square bracket is certainly negative for $x \in [0,2] $ .
But is this legal? Can I pair terms like this? Why? I am sure I have seen a case where grouping terms has given different answers to the evaluation of an infinite sum. If you know of a case I am talking about could you please give it as an example. It would help me understand better if I saw when it was not allowed and when it was.
 [/INST] The thing you are talking about is called rearrangement of terms of a series. For example
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} + \dots = \sum_{k=1}^\infty (-1)^{k+1} \frac{1}{k} = \log2 \text{.}
$$
On the other hand
$$
1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\dots=\\
(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}+\dots=\\
\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12} + \dots = \frac{1}{2} \log 2.$$
Now the operation you have made with your series does not change its arrangement, so it does not change its sum. Hope that helps.
</s>