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,
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.