Suppose we have a finite number of terms for Taylor expansion of a conditionally convergent function.
For example, $f=\frac1{1-x}$ with expansion $f=\sum_{n=0}^\infty x^n$. This expansion diverges for $|x|\ge1$. On the other hand, having infinite number of series terms, we could re-expand it at a point farther from the pole and do it repeatedly, thus finding a sequence of analytical continuations to any desired point.
The question is: can we do something similar, but with finite number of Taylor coefficients? I.e. can we approximate $f$ for $x:|x|\ge1$ using finite number of Taylor coefficients of expansion at $x=0$?
Best Answer
No. Once you restricted yourself to a finite number of terms, you have a polynomial. A polynomial $p(z)=\sum_{n=0}^d c_n z^n$ can be expanded in terms of $(z-a)$ for any point $a$ (just by using the binomial formula), but this does not change the polynomial: you still have the same values $p(z)$ at every $z$, just written differently. So, if the polynomial $p$ did not do a good job of approximating $f$ around $a$, the algebraic manipulations won't change that.
The magic of analytic continuation is possible because reshuffling pieces of an infinite series can change its behavior.