Application of the Comparison Test from Calculus I Background: My students in Calculus I have been given the task of determining the interval of convergence for the Taylor series of $f(x)=x^{0.2}$ centered at $a=7$. Writing out a few terms, it is easy to see we have $$7^{0.2}+\sum_{n=1}^\infty\frac{(-0.8)\cdots(1.2-n)}{n!}7^{0.2-n}(x-7)^n=7^{0.2}-\sum_{n=1}^\infty \frac{\prod_{j=1}^{n-1}(5j-1)}{5^nn!}7^{0.2-n}(7-x)^{n}.$$Now, it is straightforward to check that it has a radius of convergence $R=7$; the problem I'm having is checking the endpoints. Checking $x=14$ is, again, straightforward, but $x=0$ has given me trouble.
Wolfram Alpha says that the series converges using the comparison test and Maple gives the value of $0$ for the whole expansion (which makes sense since $f(0)=0^{0.2}=0$).
Are there any slick comparisons one can use to prove convergence of this series?
 A: I am surprised anything was straightforward :) This is the sort of situation where the ratio test fails and you need to use Stirling's formula to see what's going on. To save my brain, I'll do the corresponding example with $f(x)=x^{1/2}$ and $a=1$. Then we end up having to analyze the convergence of the series
$$\sum \frac{1\cdot 3\cdot \ldots \cdot (2k-1)}{2^{k+1} (k+1)!} = \sum \frac{(2k)!}{2^{2k+1}(k+1)(k!)^2}\,.$$
By the refined form of Stirling's formula, $n!\sim \left(\frac ne\right)^n\sqrt{2\pi n}$, and you can check that the terms approach $1/(2\sqrt{\pi k})$, and so the series diverges at $x=0$. This is subtle stuff, indeed. (I wouldn't be shocked if I made an error somewhere in here; corrections welcome.)
EDIT: Embarrassingly, when I worked this out, I dropped the $k+1$ term in the denominator. So the correct asymptotic estimate is $$\frac1{2(k+1)\sqrt{\pi k}}\,,$$
and by limit comparison with $\sum k^{-3/2}$, the series does, in fact, converge. My apologies. All the hard work was right; just the easy term got dropped.
A: It seems to me that your question amounts to asking about the the convergence of Newton's binomial series -- i.e., the Taylor series for $f(x) = (1+x)^{\alpha}$ centered at $0$ -- at the left endpoint of the interval of convergence.  Whether one has divergence, conditional convergence or absolute convergence depends on $\alpha$.  The full result for convergence of the binomial series at the endpoints of the interval of convergence is recorded as Theorem 12.7 in these notes.  In particular, since here $\alpha = 0.2 > 0$, we have absolute convergence at both endpoints.
I also agree that this is too advanced for Calculus I.  The notes I linked to are from a course taught at the University of Georgia called Honors Calculus With Theory, which to me at least is not really a version of "Calculus I" but rather has that as a prerequisite.  This course and the notes are instead at the level of Spivak's Calculus, which as he himself as noted, is more accurately viewed as a certain kind of undergraduate analysis text.  On the other hand I just finished teaching Calculus II, in which the single biggest topic is sequences and series.  What did I say about the binomial series?  Almost nothing.  There was a problem on the final exam asking them for the third Taylor polynomial of $f(x) = \sqrt{x}$ centered at $1$: more than that belongs in a more advanced course.
Added: Moreover, when $\alpha > 0$ the series converges to what it is supposed to, namely, $f(-1) = (1-1)^{\alpha} = 0$.  This is also proved in the notes: see Theorem 12.8.  The evaluation of the series at the endpoints is so un-straightforward that I felt the need to reference the source, which is notes of John Labute for an undergraduate analysis course at McGill University.
These results are of course well known -- I think they were known to Newton -- but are relatively hard to find in contemporary texts.  Undergraduate analysis is a bit watery these days compared to what it used to be (and graduate analysis tends to begin with much fancier things, like abstract measure theory): there are too many other mathematical fields competing for students' attention.  Too bad...
