Showing a function grows like a certain power of $n$

Question

Consider $f(n) = \frac{1}{n} {n \choose (n+1)/2} \frac{1}{2^n}$ for odd $n$. I suspect that $\sum_{n \geq 1} n^{\beta}f(n)= \infty$ for $\beta \geq 1/2$. For this purpose, it would be enough to show that $f(n) \sim n^{-3/2}$. How can I show this last statement?

Answer 1

I would use Stirling's Formula $$\binom{n}{k}\approx \frac{e^k}{\sqrt{2\pi k}}\times \left({\frac{n}{k}-\frac{1}{2}}\right)^k.$$ Then $$\binom{n}{(n+1)/2}\approx \frac{e^{(n+1)/2}}{\sqrt{\pi (n+1) }}\left( \frac{3n-1}{n+1}\right)^{(n+1)/2}.$$

Thus $$\frac{1}{2^nn}\binom{n}{(n+1)/2}\approx \frac{1}{n^{3/2}}\times\sqrt{\frac{2}{\pi}}$$ Perhaps this can help.

Answer 2

Convert the polynomial in the representation with Gamma function $$ \eqalign{ & b(n) = \left( \matrix{ n \cr n/2 + 1/2 \cr} \right) = {{\Gamma \left( {n + 1} \right)} \over {\Gamma \left( {n/2 + 3/2} \right)\Gamma \left( {n/2 + 1/2} \right)}} = \cr & = {{\Gamma \left( {n + 1} \right)} \over {\left( {n/2 + 1/2} \right)\Gamma \left( {n/2 + 1/2} \right)^{\,2} }} = {{\Gamma \left( {2\left( {n/2 + 1/2} \right)} \right)} \over {\left( {n/2 + 1/2} \right)\Gamma \left( {n/2 + 1/2} \right)^{\,2} }} \cr} $$

Apply the duplication formula $$ \Gamma \left( {2\,z} \right) = {{2^{\,2\,z - 1} } \over {\sqrt \pi }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right) $$ to get $$ b(n) = {{{{2^{\,n} } \over {\sqrt \pi }}\Gamma \left( {n/2 + 1} \right)} \over {\left( {n/2 + 1/2} \right)\Gamma \left( {n/2 + 1/2} \right)}} = {{2^{\,n} } \over {\left( {n/2 + 1/2} \right)\sqrt \pi }}{{\Gamma \left( {n/2 + 1} \right)} \over {\Gamma \left( {n/2 + 1/2} \right)}} $$

Now apply the Stirling approximation $$ \Gamma (z) \approx \sqrt {\,{{2\,\pi } \over z}\,} \left( {{z \over e}} \right)^{\,z} $$ to arrive to $$ \eqalign{ & b(n) \approx {{2^{\,n} } \over {\left( {n/2 + 1/2} \right)\sqrt {\pi \,e} }}\sqrt {\,{{n/2 + 1/2} \over {n/2 + 1}}\,} {{\left( {n/2 + 1} \right)^{\,n/2 + 1} } \over {\left( {n/2 + 1/2} \right)^{\,n/2 + 1/2} }} \approx \cr & \approx {{2^{\,n} } \over {\sqrt {\left( {n/2 + 1/2} \right)} \sqrt {\pi \,e} }}\left( {{{n/2 + 1} \over {n/2 + 1/2}}} \right)^{\,\left( {n + 1} \right)/2} \approx \cr & \approx {{2^{\,n + 1/2} } \over {\sqrt {\left( {n + 1} \right)} \sqrt {\pi \,e} }}\left( {1 + {1 \over {n + 1}}} \right)^{\,\left( {n + 1} \right)/2} \approx \cr & \approx {{2^{\,n + 1/2} } \over {\sqrt {\left( {n + 1} \right)} \sqrt \pi }} \cr} $$

Inserting the other terms, finally you get $$ f(n) = {1 \over {n\,2^{\,\,n} }}b(n) \approx \sqrt {{2 \over \pi }} {1 \over {n\sqrt {n + 1} }} \approx \sqrt {{2 \over \pi }} \;n^{\, - 3/2} $$

Answer 3

Letting $n=2m+1, m\in\mathbb{N}$ we obtain \begin{align*} \color{blue}{f(2m+1)}&=\frac{1}{2^{2m+1}(2m+1)}\binom{2m+1}{m+1}\\ &=\frac{1}{2^{2m+1}(m+1)}\binom{2m}{m}\tag{1}\\ &\,\,\color{blue}{=\frac{1}{2^{2m+1}}C_m}\tag{2} \end{align*} with $C_m$ the $m$-th Catalan-number.

In (1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

The asymptotic behaviour of the Catalan numbers $C_m$ can be shown using Stirling's formula and is according to this MSE answer

\begin{align*} C_m\sim\frac{4^m}{\sqrt{\pi}m^{\frac{3}{2}}} \end{align*}

We conclude together with (2) \begin{align*} \color{blue}{f(2m+1)\sim \frac{1}{2\sqrt{\pi}m^{\frac{3}{2}}}} \end{align*}

Note: Since $f(n)$ is given for odd $n$ only, we do not know anything about the behaviour of $f$ for even arguments.

Answer 4

In fact $f_n\sim \dfrac{1}{n}$. We know that $\dfrac{\binom{n}{\frac{n+1}{2}}}{2^n}$ is the probability of drawing $\dfrac{n+1}{2}$ heads or tails in $n$ independent coin tosses. According to WLLN (Weak Law of Large Numbers) this fraction goes to $1$ since such an event (drawing $\dfrac{n+1}{2}$ heads or tails in $n$ independent coin tosses) is the typical set. Therefore $f_n\sim\dfrac{1}{n}$ and the series diverges.