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**1. Prove that the MacLaurin series for cosx converges to cosx for all x.**

**2. Relevant equations**

Ʃ(n=0 to infinity) ((-1)^n)(x^2n)/((2n)!) is the MacLaurin series for cosx

|Rn(x)|[itex]\leq[/itex]M*(|x|^(n+1))/((n+1)!) if |f^(n+1)(x)|[itex]\leq[/itex]M

lim(n->infinity)Rn=0 then a function is equal to its Taylor series within the interval of convergence.

**3. The attempt at a solution. I already know what the solution is but I can't figure out why we are using this solution.**

First you get: |f^(n+1)(x)|[itex]\leq[/itex]1 because |cosx| and |sinx| functions are always bounded by 1

Then you write |Rn(x)|[itex]\leq[/itex](|x|^(n+1))/((n+1)!)

lim as n->infinity of (|x|^(n+1))/((n+1)!)=0, therefore lim as n-> infinity of |Rn|=0 so the MacLaurin series of cosx converges to cosx for all x.

My question

Why don't we just write Rn = [itex]\Sigma[/itex] from (i=n+1 to infinity) ((-1)^i)*(x^2i)/((2i)!) and we can see from this that the limit as n->infinity makes Rn go to zero, therefore cosx converges to cosx? Why did we have to go through all of that other stuff?

**1. The problem statement, all variables and given/known data**

**2. Relevant equations**

**3. The attempt at a solution**