calculus - Evaluate limit of function through its series expansion

The MacLaurin series expansion of the function

$$f(x) = \frac{1}{1 - x}$$

is

$$f(x) = 1 + x^2 + x^3 + x^4 + \ldots = \sum_{i = 0}^{+\infty} x^i, \ (x \neq 0)$$

so all the powers of $x$ taken with a positive sign.

We have

$$\lim_{x \to 1^-} \frac{1}{1 - x} = + \infty$$

while

$$\lim_{x \to 1^+} \frac{1}{1 - x} = - \infty$$

The first limit can be easily predicted from the series expansion: when $x$ is slightly greater than $1$, all the terms sum: the final result of their sum should be $+\infty$. The series expansion becomes in fact

$$f(x)_{x = 1} = \left( \sum_{i = 0}^{+\infty} x^i \right)_{x = 1} = \lim_{i \to +\infty} i \cdot 1 = +\infty$$

But this is true for both $x \to 1^+$ and $x \to 1^-$ and it shouldn't!

1) How can the limit for $x \to 1^+$ be correctly predicted from the series expansion?

The above series is centered in $x = 0$; $x = 1$ is quite far away from that point; but with more terms included, the series should provide the exact function values over a wider range, so even $x = 1$. But the result, even with more terms, is the same: the series limit is always $+\infty$.

2) Is this a correct perspective, or the problem should be dealt with in another way?

Answer

The issue here is that the series expansion isn't valid for $x \geq 1$. Remember that any infinite series has a radius of convergence $R$; that is, $R$ is the value such that
$$f(x) = \sum_{n=0}^\infty a_nx^n$$
converges for $|x| < R$. In the case of the geometric series that you state, we have that $R = 1$. So attempting to evaluate the series for $x > 1$ doesn't make sense.

The point is that the equality
$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$
isn't always true. It only works as long as $|x| < 1$, and so for any other values of $x$, the given statement is false.

One should note that while you get the "right" answer for $x = 1$, you should still be careful about that. In this case they agree, but it isn't always true that they will (e.g. evaluating both sides at $x = -1$ yields something false).