Problem regarding infinite sum of remainders.

Before here @math.SE there was a question regarding a problem on a maths magazine. I decided to look at the link provided, and one problem proposed was (if I'm not recalling this wrongly):

Find $$\sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\left[ {f\left( x \right) - {T_n}\left( x \right)} \right]}$$

I'm most certain it was from the MMA.

Is this the correct wording of the problem? If so, any hints on how to solve it? I'm thinking of using the integral form of the remainder, but since I don't really know what the problem really was, I can't move on.

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What are $f$ and $T_n$? –  Brian M. Scott May 10 '12 at 3:09
@BrianM.Scott A probably not so arbitrary function $f$ and its Taylor polynomial approximation. –  Pedro Tamaroff May 10 '12 at 3:10
Yes, I could (with some thought) guess that, but it really ought to be in the problem statement. –  Brian M. Scott May 10 '12 at 3:10
@BrianM.Scott Maybe it was AMS or Mathematical Monthly –  Pedro Tamaroff May 10 '12 at 3:12

I am assuming that $$T_n(x)=\sum_{k=0}^nf^{(k)}(0)\frac{x^k}{k!}\tag{1}$$ Then, for $x$ inside the radius of convergence of $f(x)$ about $x=0$, \begin{align} \sum_{n=0}^\infty(-1)^n[f\left(x\right)-T_n(x)] &=\sum_{n=0}^\infty(-1)^n\sum_{k=n+1}^\infty f^{(k)}(0)\frac{x^k}{k!}\\ &=\sum_{k=1}^\infty\sum_{n=0}^{k-1}(-1)^n f^{(k)}(0)\frac{x^k}{k!}\\ &=\sum_{k=1}^\infty\frac{1-(-1)^k}{2}f^{(k)}(0)\frac{x^k}{k!}\\ &=\frac12\left(\sum_{k=0}^\infty f^{(k)}(0)\frac{x^k}{k!}-f^{(k)}(0)\frac{(-x)^k}{k!}\right)\\ &=\frac{f(x)-f(-x)}{2}\tag{2} \end{align} which is the odd part of $f$.
Prompted by a question from Peter Tamaroff, let's consider what would happen if the Taylor series were taken about $x=a$. Following the same logic as above we would get \begin{align} \sum_{n=0}^\infty(-1)^n[f\left(x\right)-T_n(x)] &=\frac12\left(\sum_{k=0}^\infty f^{(k)}(a)\frac{(x-a)^k}{k!}-f^{(k)}(a)\frac{(a-x)^k}{k!}\right)\\ &=\frac12\left(\sum_{k=0}^\infty f^{(k)}(a)\frac{(x-a)^k}{k!}-f^{(k)}(a)\frac{((2a-x)-a)^k}{k!}\right)\\ &=\frac{f(x)-f(2a-x)}{2}\tag{3} \end{align}
Would the solution be similar if the Taylor series was around $a$? –  Pedro Tamaroff May 10 '12 at 3:35
@Peter: If the Taylor series were about $x=a$, then we would get $\dfrac{f(x)-f(2a-x)}{2}$. –  robjohn May 10 '12 at 3:51
Assuming $f(x) = \sum_{k=0}^\infty c_k x^k$ is an analytic function and $T_n(x) = \sum_{k=0}^n c_k x^k$ is the Maclaurin polynomial of degree $n$, and $|x|$ is less than the radius of convergence, we have a bound $|c_k x^k| < A r^k$ where $r < 1$. The series then converges absolutely. Now $c_k x^k$ occurs in $(-1)^n (f(x) - T_n(x))$ with coefficient $(-1)^n$ if and only if $k > n$. Since $\sum_{n=0}^{k-1} (-1)^n = 1$ if $k$ is odd and $0$ if $k$ is even, the answer is $\sum_{k\ \text{odd}} c_k x^k = (f(x) - f(-x))/2$.