Prove convergence of MacLaurin expansion of$ (1+x)^{1/2}$ at $|x|=1$.

Question

I know that the Maclaurin expansion of $(1+x)^{1/2}$ has a domain of convergence of $|x|\leq1$. It is easy to prove using the ratio test that this expansion is convergent when $|x|<1$, however, I don't know how to prove convergence when $|x|=1$. Does anybody know any simple way to prove this convergence?

Answer 1

If you carefully write it out, you will get (for $c$ either $3/2$ or $1/2$):

$$f(\pm1)\,-\,c\,\,=\,\,-\frac{2^{-2}}{1(2)}\,\pm\,\frac{2^{-3}(3)}{1(2)(3)}\,-\frac{2^{-4}(3)(5)}{1(2)(3)(4)}\,\pm\,\frac{2^{-5}(3)(5)(7)}{1(2)(3)(4)(5)}\,-\pm\dots.$$

We want to show absolute convergence of this series for either choice $x=+1$ or $x=-1$, so make every term positive and bring each $2^{-k}$ factor into the denominator:

$$\sum_{k\,\geq\,2} \frac{(2k-3)!!}{(2k)!!} \,=\, \sum_{k\,\geq\,2} \frac{({\sqrt{2k-3}})^2!!}{(2k)!!}.$$

The general term of this series we can bound above by the term of the harmonic series of exponent $p=3/2$, i.e. over-harmonic series or $p$-series with $p>1$. This via the estimate $\sqrt{n+1}\sqrt{n-1}\leq n$:

$$\frac{\sqrt{1}}{4(2)}\,+\,\frac{\sqrt{3}\sqrt{3}\sqrt{1}}{6(4)(2)}\,+\,\frac{\sqrt{5}\sqrt{5}\sqrt{3}\sqrt{3}\sqrt{1}}{8(6)(4)(2)}\,+\,\frac{\sqrt{7}\sqrt{7}\sqrt{5}\sqrt{5}\sqrt{3}\sqrt{3}\sqrt{1}}{10(8)(6)(4)(2)}\,+\dots$$

$$=\frac{\sqrt{1}}{4(2)}\,+\,\frac{\sqrt{3}}{6(4)}\,\frac{\sqrt{3}\sqrt{1}}{2}\,+\,\frac{\sqrt{5}}{8(6)}\frac{\sqrt{5}\sqrt{3}}{4}\frac{\sqrt{3}\sqrt{1}}{2}\,+\dots$$

$$\leq\frac{\sqrt{1}}{4(2)}\,+\,\frac{\sqrt{3}}{6(4)} \,+\,\frac{\sqrt{5}}{8(6)} \,+\,\frac{\sqrt{7}}{10(8)}\,+\dots=\sum_{k\,\geq\,2} \frac{(2k-3)^{\frac{1}{2}}}{(2k)(2k-2)}.$$

Answer 2

A more general result states that $$ (1+x)^\alpha, $$ where $\alpha\in\mathbb C$ with $\Re(\alpha)>0$, converges absolutely at $|x|=1$.

The proof as well as a more detailed description of the convergence issues can be found here.