<p>i guess you would have to find a function f whose Taylor polinom equals your sum. </p>

<p>In your case (i hope i'm not making some huge mistake :D) :

we use S[ ]as sigma to denote a sumation.

f'{k}(x) = the k derivate of f(x)</p>

<p>f(x)=lim(k->infinitiy) f(0)+S[f'{k}(0)*x^k/k!] </p>

<p>i hope that is right </p>

<p>f(x^4)=lim(k->infinitiy) f(0)+S[f'{k}(0)*x^(4k)/k!]

right?</p>

<p>so if we simply find a function that gives us f'{k}(0)=(-1)^k then we should have the sum</p>

<p>say e^(-x). </p>

<p>So what is the McLaurin polynom of f(x^4)?</p>

<p>f(x^4)=f(0)+f'(0)<em>x^4+f''(0)</em>x^8/2!+....</p>

<p>so your sumation:

summation from n=0 to infinity of [(-1)^n * x^(4n)] / n! = e^(-x^4)</p>

<p>Hope i'm right and i won't end up being the laughing stock of anyone who knows math :p</p>