7
$\begingroup$

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

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)

  • 0
    The proofs in question id 9372 show that the above series converges at $1$ and thus converge for $|x| \lt 1$ (radius of convergence).2012-04-26

1 Answers 1

2

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 $|x|\leqslant1$.