Edit: I changed the inequality to the one that I think was meant to be asked. This is a former exam question from my math dept, and it is relatively old - from 1993. So, I think there was a typo on the R.H.S. of the inequality. Thanks,

I am trying to prove that, for $x>0$

$$\big(1 + \frac{1}{2}x - \frac{1}{8}x^2\big) < \sqrt{1+x} < \big(1+ \frac{1}{2}x\big)$$

It's easy to notice that the left term is just the Taylor expansion of the middle term, using the generalized Binomial series.

I don't know how to achieve the inequality, so I expanded out to a few more terms to see whether I can say something more:

$$\sqrt{1+x}= \big(1 + \frac{1}{2}x - \frac{1}{8}x^2 +\frac{1}{16}x^3-\frac{5}{128}x^4 + ... \big)$$

So now it's more obvious that $\sqrt{1+x}$ has an alternating Taylor series, valid for $|x|<1$. Also, the coefficients are monotone decreasing to zero, in absolute value.

Where can I go from here?

Any ideas are welcome.


  • 2
    $\begingroup$ There is at least one error in your inequalities. How can $\sqrt{1+x}$ be bounded above by $3/2$? Please clarify. $\endgroup$ – gammatester Dec 14 '15 at 8:48
  • 1
    $\begingroup$ I think he meant $1+{1\over2}x$. $\endgroup$ – cr001 Dec 14 '15 at 8:53
  • $\begingroup$ Hi @gammatester, perhaps the right hand term is $(1+\frac{1}{2} x)$. $\endgroup$ – User001 Dec 14 '15 at 8:53
  • $\begingroup$ I think so too @cr001 -- but as I read it here on paper, it is 1/2. But also, this is a 1993 exam question, so there's definitely a possibility of transcription error. $\endgroup$ – User001 Dec 14 '15 at 8:54
  • 1
    $\begingroup$ The right inequality is Bernoulli's IE, it is strict because because $x>0$. $\endgroup$ – gammatester Dec 14 '15 at 9:02

If $x>0$ we have that $\left(1+\frac{x}{2}\right)^2 = 1+x+\frac{x^2}{4} > 1+x$, hence $1+\frac{x}{2}>\sqrt{1+x}$.

On the other hand, $$1+\frac{x}{2}-\sqrt{1+x} = \frac{\left(1+\frac{x}{2}\right)^2-(1+x)}{1+\frac{x}{2}+\sqrt{1+x}}<\frac{\frac{x^2}{4}}{2}$$ gives the other side of the wanted inequality.

  • 1
    $\begingroup$ Thanks for such a pretty solution @JackD'Aurizio. I had expected a nasty Taylor estimate, so this is a nice surprise. Have a great night :-) $\endgroup$ – User001 Dec 14 '15 at 9:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.