Find $\lim_{n\to\infty} \sqrt[n]{a_n}$ where $a_n=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n-k}{k}$

Question

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...

Answer 1

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 $$\boxed{\lim_{n\to\infty}\sqrt[n]{a_n}=\frac{\sqrt{5}+1}2}$$

Answer 2

Clearly,

$$ \max_{0\le k\le\frac{n}{2} }C_k^{n-k}\le a_n=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} C_k^{n-k}\le \frac{n}{2}\max_{0\le k\le\frac{n}{2} }C_k^{n-k} $$ Hence if we show that $$\left(\max_{0\le k\le\frac{n}{2} }C_k^{n-k}\right)^{1/n}\to L,$$ then we shall have that $a_n^{1/n}\to L$.

Next observe that $$ C_k^{n-k}=\frac{(n-2k)(n-2k+1)}{k(n-k)}\cdot C_{k-1}^{n-k+1} $$ and hence $C_k^{n-k}$ is maximised (for fixed $n$) asymptotically when $k=\dfrac{(5-\sqrt{5})n}{10}$. So $$ \max_{0\le k\le\frac{n}{2} }C_k^{n-k}\approx \frac{((1-c)n)!}{(cn)!((1-2c)n)!}, \quad c=\dfrac{5-\sqrt{5}}{10}. $$ Next we use that fact that $\dfrac{(n!)^{1/n}}{n}\to\dfrac{1}{\mathrm{e}}$ Hence $$ \left(\frac{((1-c)n)!}{(cn)!((1-2c)n)!}\right)^{1/n}= \left(\frac{\frac{((1-c)n)!}{((1-c)n)^{(1-c)n}}}{\frac{(cn)!}{(cn)^{cn}}\frac{((1-2c)n)!}{((1-2c)n)^{(1-2c)n}}}\right)^{1/n}\cdot \frac{(1-c)^{1-c}}{c^c(1-2c)^{1-2c}}\to \frac{(1-c)^{1-c}}{c^c(1-2c)^{1-2c}}. $$

Answer 3

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