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

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 $n\to\infty$

• Would be nice to show us the expression that you established.
– user65203
Jun 9, 2016 at 18:05
• Are you sure you've written out those repeated products right?
– user14972
Jun 9, 2016 at 19:13

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 $n \to \infty$.

• $c$ is some unknown point between $x$ and $0$, how can I control it to be small? I need both $|x|<1$ and $|1+c|<1$. Jun 10, 2016 at 11:36
• @Joshhh You must have $|x|<1$ or the series doesn't converge absolutely. Jun 10, 2016 at 21:39
• How do you show that $(2n-3)!! < (n+1)!$? I have failed to do this by induction Jun 13, 2016 at 16:48
• When I put this expression in Mathematica, it seems that the ratio between the two diverge, i.e. from some point $(2n-3)!! > (n+1)!$ Jun 13, 2016 at 17:06
• Ah, it seems you're right, so this answer is incorrect. Jun 13, 2016 at 18:01