As title, I am interested in the value of $$\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{n+k+k^\alpha},$$ where $\alpha\in \mathbb R$. Here is my attempt:
- For $\alpha\in (0,1)$, note that $$\sum_{k=1}^n\left(\frac{1}{n+k}-\frac{1}{n+k+k^\alpha}\right)=\sum_{k=1}^n\frac{k^\alpha}{(n+k+k^\alpha)(n+k)}\leq \sum_{k=1}^n \frac{n^\alpha}{n^2}=\frac{1}{n^{1-\alpha}}\to 0.$$ Thus, we can substitute $1/(n+k+k^\alpha)$ by $1/(n+k)$ in the summation. Then by standard Riemann sum, we know that $$\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{n+k+k^\alpha}=\int_0^1 \frac{1}{1+x}dx=\ln(2).$$
- For $\alpha\leq 0$, it follows that $$\sum_{k=1}^n \frac{1}{n+k+\sqrt{k}}\leq \sum_{k=1}^n \frac{1}{n+k+k^\alpha}\leq \sum_{k=1}^n \frac{1}{n+k}.$$ Then the result follows by sandwich theorem and the previous case.
However, I don't know whether there is any approach to deal with the case $\alpha>1$ (for $\alpha=1$, it is easy to show that the value is given by $\ln\sqrt{3}$). I personally "believe" that $$\lim_{n\to \infty}\sum_{k=1}^n \left(\frac{1}{n+k^\alpha}-\frac{1}{n+k+k^\alpha}\right)=0,$$ yet as mentioned above, I couldn't find a way to verify this result.