Please help. The answer is in the brackets, but how to get there...
View attachment 21654
Here are a couple of approximations that may help you sometimes:
\(\displaystyle \sqrt{a + b} \ \approx \ \sqrt{a} + \dfrac{1}{2}b\)
\(\displaystyle \sqrt[3]{a + b} \ \approx \ \sqrt[3]{a} + \dfrac{1}{3}b\)
For this problem, a = 1 and b = 2/n for the cube root approximation.
The expression then is approximated by
\(\displaystyle \dfrac{n}{2}\bigg(1 + \dfrac{2}{3n} \ - \ 1 \bigg)\)
Simplify that and then take the limit as n approaches infinity.