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.


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 $n\to\infty$

  • $\begingroup$ Would be nice to show us the expression that you established. $\endgroup$
    – user65203
    Jun 9, 2016 at 18:05
  • 1
    $\begingroup$ Are you sure you've written out those repeated products right? $\endgroup$
    – user14972
    Jun 9, 2016 at 19:13

1 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 $n \to \infty$.

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

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