Find $\lim_{n\to\infty} \sqrt[n]{a_n}$ where $a_n=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n-k}{k}$ Let $a_n=\sum_{k=0}^{[n/2]} C_{n-k}^k$. Find $\lim_{n\to\infty} \sqrt[n]{a_n}$. Here $[n/2]$ is the largest integer $\leq n/2$, $C_{n-k}^k=\frac{(n-k)!}{k!(n-2k)!}$.

It sounds $a_{2m+1}-a_{2m}=a_{2m-1}$. Then...
We have the following lemma which follows from a stronger inequality proven in this question:


  If $(a_n)$ is a sequence in $\mathbb{R}^+$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ exists then
  $$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$


For this sequence we have that
$$a_{2n+1}=a_{2n}+a_{2n-1}$$
$$\frac{a_{2n+1}}{a_{2n}}=1+\frac{a_{2n-1}}{a_{2n}}\tag{1}$$
Also, by taking $n\to\infty$, we have that 
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{a_{2n+1}}{a_{2n}}=\lim_{n\to\infty}\frac{a_{2n}}{a_{2n-1}}\tag{2}$$
Hence if we denote the desired limit by
$$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L$$
then, using $(1)$ and $(2)$, we get the equation
$$L=1+\frac1{L}$$
which we can solve for $L\gt0$ giving
$$\boxed{\lim_{n\to\infty}\sqrt[n]{a_n}=\frac{\sqrt{5}+1}2}$$
How about schematically:

$a_n =$ the sum of the first half of the nth row of pascal's triangle

= half of 2^n when n is odd and a bit more than that when n is even

= roughly $2^{(n-1)}$

Hence the nth root of this is $2^{(1-1/n)}$ which tends to 2 as n-> inf