Need help performing the taylor series expansion for below-mentioned expression in square brackets

Everyone seems to have overlooked that in a Taylor series, the variable appears in the numerator, not the denominator. What you have is not the Taylor series of the function with respect to $y$, but instead with respect to $\frac 1y$.

That is, if you let $x = \frac 1y$, the function becomes $$f(x) = 1-\sqrt{1+x}$$ The Taylor series for this function begins with $$f(x) = 0 - \frac12 x + \frac 18x^2 + \dots$$ Dropping the higher order terms and substituting $\frac 1y$ for $x$, we get

$$f\left(\frac 1y\right) \approx -\frac 1{2y}$$

Which is the result you should be after. The negation is correct, your version is not. For positive $y$, the function has a negative value.