How to find the limit of $\lim_{n\rightarrow\infty}\frac{[\sum_{i=0}^{n} (2i+1)^\alpha]^{\beta+1}}{[\sum_{i=1}^{n} (2i)^\beta]^{\alpha+1}}$?

Question

How to find this limit ： $$\lim_{n\rightarrow\infty}\frac{[\sum_{i=0}^{n} (2i+1)^\alpha]^{\beta+1}}{[\sum_{i=1}^{n} (2i)^\beta]^{\alpha+1}}$$ where $\alpha \;and\; \beta \;$are real numbers.

Answer 1

Now I get the answer: if $ \alpha\neq-1~and~\beta\neq-1$: $$\lim_{n\rightarrow\infty}\frac{[\sum_{i=0}^{n} (2i+1)^\alpha]^{\beta+1}}{[\sum_{i=1}^{n} (2i)^\beta]^{\alpha+1}}\\= 2^{\alpha-\beta}\lim_{n\rightarrow\infty}\frac{\{\frac{2}{n}[\sum_{i=0}^{n}(\frac{2i+1}{n})^\alpha]\}^{\beta+1}}{\{\frac{2}{n}[\sum_{i=1}^{n}(\frac{2i}{n})^\beta]\}^{\alpha+1} }\\ =2^{\alpha-\beta}\lim_{n\rightarrow\infty}\frac{(\int_{0}^{2}t^\alpha{\rm{d}}t+(\frac{2n+1}{n})^\alpha\frac{2}{n})^{\beta+1}}{(\int_{0}^{2}t^\beta{\rm{d}}t)^{\alpha+1}}\\=2^{\alpha-\beta}\frac{(\beta+1)^{\alpha+1}}{(\alpha+1)^{\beta+1}} $$And it's easy if $\alpha=-1$ or $\beta=-1$.