The Power of Taylor Series

Not quite on the power of Taylor's series, but you could use the Taylor series for $f(x) = e^x$ to show that $e$ is irrational.

Here is an interesting application of power series; unfortunately one would need to bother with the remainder to make it really interesting.

$$\arctan(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1} \,.$$

Plug in $x= \frac{1}{\sqrt 3}$ ($1$ would also work but one would need to explain why this formula also holds at the end point of the interval). We get:

$$\frac{\pi}{6} = \frac{1}{\sqrt{3}}\sum_{n=0}^\infty \frac{(-1)^n}{3^n2n+1} \,.$$

The right side is an alternating series which converges very fast, thus you can use it to calculate $\pi$ with 5-6 digits. And it is alternating, which means you could use the Alternating Series error estimate.

You can also do the same for the Taylor series of $e^x$.

I presume the students would know how to sum (finite/infinite) geometric series. You can tell them that the infinite geometric series can be seen as the Taylor expansion of the function $\frac{1}{1-x}$: $$ \frac{1}{1-x} = 1+ x + x^2 + x^3 + \cdots. $$ Integrating term by term (not always valid!), $$ -\ln (1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots, $$ which is nothing but the Taylor expansion of the function $\ln(1+x)$. (Of course, you can also derive the Taylor expansion directly, if you prefer.) Finally, plugging in $x=-1$ (once again, not really valid!), we get the sum of the alternating harmonic series: $$ \ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots. $$

There are many unjustified steps in the above derivation, but it gives a cool demonstration of power series and Taylor expansions.

You can show that a certain function $f$ does not have any zero between $-1$ and $1$, for instance, by showing that $1/f$ is analytic with a radius of analyticity at least $1$ (and therefore finite).