Prove $\sqrt{1+x}$ can be represented by a power series

Question

I need to show that $\sqrt{1+x}$ can be represented as a power series. I need to prove the equality between the function and its Taylor series, not to prove that the Taylor series of the function, $\sum\limits_{n=0}^{\infty}{\frac{1}{2}\choose n}x^n$ converges in $(-1,1)$.

My lead is to use the Lagrange form of the remainder, by I do not know how to show the remainder tends to zero.

Edit:

The $n$-th derivative is $$ f^{(n)}(0) = \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdot\ldots\cdot\left(\frac{1}{2} - n + 1\right)}{n!} $$ and hence the Lagrange form of the remainder is $$ R_n(x) = \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdot\ldots\cdot\left(\frac{1}{2} - n + 2\right)}{(n+1)!}(1+c)^{\frac{-1-2n}{2}}x^n $$

But I do not know how to show this tends to $0$ as $n\to\infty$

Answer 1

Start by simplifying the expression for $R_n(x)$. The large fraction becomes, after taking absolute values:

$$ \frac{1}{(n+1)!} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdots \frac{2n - 3}{2} = \frac{(2n - 3)(2n - 5) \cdots (3)(1)}{2^{n+1}(n+1)!} $$

Observe that the numerator is bounded above by $(n + 1)!$, and so the large fraction is smaller than $2^{-n-1}$. So we have:

$$ |R_n(x)| \le \frac{(1+c)^{ \frac{1-2n}{2} } } {2^{n+1}} x^n = \frac{(1+c)^{1/2}}{2} \left( \frac{x}{2(1+c)} \right) ^n $$

For any $c$ between $0$ and $x$, the right hand side tends to $0$ as $n \to \infty$.