Inspired by Rohit Ner

First we prove the following identity which we shall use later, using the formula for the sum of terms in a geometric progression: $\begin{aligned} \sum_{k=0}^{2n} (1-t)^{2n}\left(\frac{t}{t-1}\right)^k &=(1-t)^{2n} \frac{(1-\left(\frac{t}{t-1}\right)^{2n+1})}{1-\frac{t}{t-1}} \\ &=(1-t)^{2n+1}+t^{2n+1} \end{aligned}$ (the last step involved splitting the numerator and taking out the $-1$ in the second term, while simplifying the denominator). Now we utilise Euler's well-known Beta function: $B(x,y)=\frac{(x-1)!(y-1)!)}{(x+y-1)!}=\int_0^1 t^{x-1}(1-t)^{y-1} dt.$ Putting what we have until here together - replacing $(x-1,y-1)$ with $(k,n-k)$ tells us $\begin{aligned} \sum_{k=0}^{2n}\frac{(-1)^k}{{2n \choose k}} &= \sum_{k=0}^{2n}\frac{(-1)^kk!(2n-k)!(2n+1)}{(2n+1)!} \\ &= \sum_{k=0}^{2n}(-1)^k(2n+1)\int_0^1 t^{k}(1-t)^{2n-k} dt\\ &= (2n+1)\int_0^1\sum_{k=0}^{2n} (1-t)^{2n}\left(\frac{t}{t-1}\right)^kdt\\ &= (2n+1)\int_0^1(1-t)^{2n+1}dt+\int_0^1t^{2n+1}dt \\ &=2(2n+1)\int_0^1t^{2n+1}dt=\frac{2n+1}{n+1} \end{aligned}$ by using the identity before to get rid of the sum. For the sake of completeness, I shall also show that the alternating sum of the harmonic series converges to $\text{ln}(2)$ . Note that $\begin{aligned} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} &= \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \\ &= \sum_{n=0}^\infty\int_0^1 (-x)^n dx\\ &=\int_0^1 \sum_{n=0}^\infty\frac{1}{x+1} dx\\ &=\text{ln}(2) \end{aligned}$ by using the representation of the sum of an infinite geometric progression which is valid as we are still within the radius of convergence. Now although Brian has presented rest of the solution, I will just complete it, we know now that the expression which limit we want to find is $\sum_{n=1}^n (-1)^{n+1}\left(2-\frac{2n+1}{n+1}\right)=\sum_{n=1}^n (-1)^{n+1}\frac{1}{n+1}$ and hence the limit is $1-\sum_{n=1}^n \frac{(-1)^{n+1}}{n}=1-\text{ln}(2),$ giving a final answer of $1^3+2^3=\boxed{9}$ .

I'll expand this solution later, but for the time being, note that

$2 - \displaystyle\sum_{k=0}^{2n} \dfrac{(-1)^{k}}{\dbinom{2n}{k}} = 2 - \dfrac{2(2n + 1)}{2n + 2} = \dfrac{1}{n + 1}.$

(See here for a discussion on this type of sum.)

Then $\lim_{N \rightarrow \infty} \displaystyle\sum_{n=1}^{N} \dfrac{(-1)^{n+1}}{n + 1} = 1 - \ln(2).$

Thus $a^{3} + b^{3} = 1^{3} + 2^{3} = \boxed{9}.$