Find the Limit!

writing the equation as $\lim_{n\rightarrow \infty} \sum_{k=0}^n \frac{\sqrt[3]{n-k}(-1)^k}{\sqrt[3]{n}}$ We know that the limit of the sum is equal to the sum of the limit (unless for ambigous limits such as $\infty-\infty$ ) therefore: $\sum_{k=0}^\infty \lim_{n\rightarrow \infty} \frac{\sqrt[3]{n-k}(-1)^k}{\sqrt[3]{n}}$ which can be reduced to: $\sum_{k=0}^\infty \lim_{n\rightarrow \infty} \sqrt[3]{1-\frac{k}{n}}(-1)^k$ and since $\lim_{n\rightarrow \infty}\frac{k}{n}=0$ we have: $\sum_{k=0}^\infty (-1)^k$ By manipulating the series we know that: $\sum_{k=0}^\infty (-1)^k=1-\sum_{k=0}^\infty (-1)^k$ that leads to
$\sum_{k=0}^\infty (-1)^k=1/2$

Why was the solution 0.5 marked incorrect when the "correct" answer is given as 0.500? Also, I wasn't given three chances!