I'm trying to solve the following limit
$$\lim_{x\to\infty} \frac{\sum_{k=0}^{x/2}\binom{x}{2k}2k(x-2k)}{\sum_{k=0}^{x/2}\binom{x}{2k}(x-2k)^{2}}$$
My thoughts:
I've tried to use Stirling's formula for the binomial coefficient $\binom{x}{2k}\approx\frac{x^{2k}}{2k!}$. Using that, the limit takes the value $-1$ which is wrong since it must be positive.
Thanks in advance.
Best Answer
It is easy to compute the numerator and the denominator explicitly. Let (for $n\geqslant 0$) $$f_n(t)=\sum_k\binom{n}{2k}t^{2k}=\frac12\big((1+t)^n+(1-t)^n\big),$$ where the summation is over all $k$ such that the term is nonzero, i.e. $0\leqslant k\leqslant\lfloor n/2\rfloor$; then $$\sum_k\binom{n}{2k}2kt^{2k}=tf_n'(t)=n\big(f_n(t)-f_{n-1}(t)\big)$$ and, using the last equality twice, $$\sum_k\binom{n}{2k}(2k)^2 t^{2k}=t\big(tf_n'(t)\big)'=n^2 f_n(t)-n(2n-1)f_{n-1}(t)+n(n-1)f_{n-2}(t).$$ Using these formulae for $\color{red}{n>2}$ and $t=1$, we get \begin{align*} \sum_k\binom{n}{2k}2k(n-2k)&=n(n-1)2^{n-3}, \\ \sum_k\binom{n}{2k}(n-2k)^2&=n(n+1)2^{n-3}. \end{align*} The answer is immediate now.