# Bounding Error term in Taylor Expansion of $\sqrt{1+x}$

I am attempting to justify the expansion

$$\sqrt{1+x}= 1 + \frac{x}{2} + \sum_{n=2}^{\infty}{(-1)^n \frac{1}{2n}\frac{(1-\frac{1}{2}) \cdots ((n-1)-\frac{1}{2})}{(n-1)!}x^n}$$

for $-1<x\leq 1$

I've got the expansion, but I cannot prove that the error term tends to zero.

$$E_n = \frac{(-1)^{n-1}(2n-3)(2n-1) \cdots (1)}{2^n n!}(1+\theta x)^{-\frac{2n-1}{2}}x^n$$

where $\theta \in (0,1)$

The question suggests using the Constancy Lemma (if the differential is zero, the function is constant), but I can't make that work either. Any help gratefully appreciated.

(While technically this is homework I'm the tutor so I allowed to cheat! Also it is extra embarrassing that I cannot do it)

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Have you tried to apply some convergence test to the remainder? It is a power series, so it must have a radius of convergence. – Siminore Apr 26 '12 at 9:32
I can't see how I would do that without expanding $(1+ \theta x)^{-\frac{2n-1}{2}}$, which would require assuming the answer of the initial question - the point is to do this without using the Binomial Theorem. – Amy Apr 26 '12 at 10:10
Well, if the Taylor series converges, then it must be THE Taylor series of your function. I mean: if the Taylor coefficients are $\{a_n\}$ and $\sum_n a_n x^n$ converges in some interval $-\varepsilon < x < \varepsilon$, then $\sum_n a_n x^n$ must be the Taylor series of your function. – Siminore Apr 26 '12 at 11:06
I'm not sure if this helps, but I think you use the double factorial to rewrite $\frac{(2n-1)(2n-3)\cdots 1}{2^n}=\frac{(2n-1)!!}{2^n}=\Gamma(n+1/2)\sqrt{\pi}$. Then you'll have $E_n=(-1)^{n-1}\sqrt{\pi}\frac{(n-1/2)!}{n!}\frac{x^n}{\sqrt{(1+\theta x)^{2n-1}}}$. – draks ... Apr 26 '12 at 12:30
possible duplicate of Why Doesn't This Series Converge? – Aryabhata Apr 26 '12 at 15:33

Consider the absolute value $$a_n=\frac{1}{2n}\frac{(1-\frac{1}{2}) \cdots ((n-1)-\frac{1}{2})}{(n-1)!}$$ of the coefficient of $x^n$ in the series expansion of $\sqrt{1+x}$. Then $$\frac{a_{n+1}}{a_n}=\frac{n-\frac12}{n+1}=1\color{red}{-\frac32}\frac1n+o\left(\frac1n\right),$$ and the heuristics in such cases is that $a_n$ behaves roughly like $n^s$ with $s=\color{red}{-\frac32}$ since, for any $s$, $$\frac{(n+1)^s}{n^s}=1+\frac{s}n+o\left(\frac1n\right).$$ To continue the proof more rigorously, choose any $t\lt\frac32$ and consider $b_n=n^ta_n$. Then $$\frac{b_{n+1}}{b_n}=\frac{(n+1)^t}{n^t}\frac{a_{n+1}}{a_n}=\left(1+\frac{t}n+o\left(\frac1n\right)\right)\cdot\left(1-\frac32\frac1n+o\left(\frac1n\right)\right),$$ hence $$\frac{b_{n+1}}{b_n}=1+\left(t-\frac32\right)\frac1n+o\left(\frac1n\right).$$ Since $t-\frac32\lt0$, this implies that $(b_n)$ is ultimately decreasing, in particular $b_n\leqslant c_t$ for every $n$, for some finite $c_t$.
Finally, $a_n\leqslant c_tn^{-t}$ uniformly over $n$ for every $t\lt\frac32$, and one can choose $t\gt1$. Then $\sum\limits_n n^{-t}$ converges (absolutely) hence $\sum\limits_na_n$ converges (absolutely) and the series expansion of $\sqrt{1+x}$ converges (absolutely) for every $|x|\leqslant1$.