I want to prove the follwing:
When $n\rightarrow \infty$ and $\alpha >2$, $$\frac{1}{n}\ln\left[2\sum_{i=0}^{n-1}\binom{\alpha n-1}{i}\right]\leq \alpha \ln(\alpha)-(\alpha-1)\ln(\alpha-1)$$ Hint: Use the Stirling approximation
I tried the following: \begin{align} \frac{1}{n}\ln\left[2\sum_{i=0}^{n-1}\binom{\alpha n-1}{i}\right]=\frac{1}{n}\ln\left[2\sum_{i=0}^{n-1}\frac{(\alpha n-1)!}{i!(\alpha n -1-i)!}\right]\\ \approx \frac{1}{n}\ln\left[2\sum_{i=0}^{n-1}\frac{\sqrt{2\pi(\alpha n -1)}(\alpha n-1)^{(\alpha n -1)}e^{-(\alpha n-1)}}{\sqrt{2 \pi i}\,i^ie^{-i} \sqrt{2\pi(\alpha n -1-i)}(\alpha n -1-i)^{(\alpha n -1-i)}e^{-(\alpha n -1-i)}}\right]\\ =\frac{1}{n}\ln\left[2\frac{(\alpha n-1)^{(\alpha n -1/2)}}{\sqrt{2\pi}}\sum_{i=0}^{n-1}\frac{1}{i^{i+1/2}(\alpha n-1-i)^{(\alpha n -i -1/2)}}\right]\\ =\frac{1}{n}\left[\ln(2)+(\alpha n -1/2)\ln(\alpha n -1)-\ln(\sqrt{2 \pi})+\ln(\sum_{i=0}^{n-1}\frac{1}{i^{i+1/2}(\alpha n-1-i)^{(\alpha n -i -1/2)}})\right] \end{align} Where in the aproximation I used the Stirling Approximation.
But now I am stucked.
Could you please help me with the sequence?