# Proving an inequality that looks related to the Binomial series,

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.

Thanks,

• There is at least one error in your inequalities. How can $\sqrt{1+x}$ be bounded above by $3/2$? Please clarify. – gammatester Dec 14 '15 at 8:48
• I think he meant $1+{1\over2}x$. – cr001 Dec 14 '15 at 8:53
• Hi @gammatester, perhaps the right hand term is $(1+\frac{1}{2} x)$. – User001 Dec 14 '15 at 8:53
• 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. – User001 Dec 14 '15 at 8:54
• The right inequality is Bernoulli's IE, it is strict because because $x>0$. – 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.