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.
 A: The terms of absolutely convergent series can be rearranged and grouped at will. The basic idea is that you can consider the subseries of positive terms and negative terms separately. Precisely because the original series converges absolutely, both subseries converge, say to $S^+$ and to $S^-$, and thus the sum of the series would be $S^++S^-$.
Detailed proofs can be easily found in most basic calculus books.
A: 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.
A: Once you know what the sum of an infinite series is defined as, it is obvious when rearrangements work and when they don't. The sum is the limit of the partial sums if it exists. So if you rearrange or group the terms in such a way that the partial sums still converge to the original limit, then the sum is unchanged. Here you have grouped the terms in pairs, so you effectively have taken alternate terms in the sequence of partial sums, which clearly must have the same limit as the original if the original converged. Thus you can always pair adjacent terms up if the original sequence converges.
