Find the value of the expression $$\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{1}{n+k^{\alpha}}\right)$$
where $\alpha$ is a positive number.
It is my first time seeing a question like this. Normally, we just put $n=\infty$ in the summand which is same as calculating the series till infinite terms.
But here it is different. The variable in the bound is also in the expression, meaning $n=\infty$ in the summand is not applicable in this case.
I expanded the series as $$\frac{1}{n+1}+\frac{1}{n+2^{\alpha}}+\cdots$$
But had no luck.
Any help is greatly appreciated.
Best Answer
Actually the person who posted the problem has also given its solution, that I would like to share.
The limit equals $1$ when $0<\alpha<1$, $\ln2$ when $\alpha=1$ and $0$ when $\alpha>1$
$$\frac{1}{n+n^{\alpha}}<\frac{1}{n+1^{\alpha}}+\frac{1}{n+2^{\alpha}}+\cdots+\frac{1}{n+n^{\alpha}}<\frac{n}{n+1}$$ and hence when $\alpha<1$ the limit equals $1$.
When $\alpha=1$ we have $$\lim_{n\rightarrow\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}=\int_0^1\frac{\operatorname{dx}}{1+x}$$ $$=\ln2$$
When $\alpha>1$ we have since $n+k^{\alpha}\ge2\sqrt{n}k^{\frac{\alpha}{2}}$ that $$0<\frac{1}{n+1^{\alpha}}+\frac{1}{n+2^{\alpha}}+\cdots+\frac{1}{n+n^{\alpha}}<\frac{1}{2\sqrt{n}}\left(\frac{1}{1^{\frac{\alpha}{2}}}+\frac{1}{2^{\frac{\alpha}{2}}}+\cdots+\frac{1}{n^{\frac{\alpha}{2}}}\right)$$ An application of Stolź-Cesaro Lemma (the $\frac{\infty}{\infty}$ case) shows that $$\lim_{n\rightarrow\infty}\frac{\frac{1}{1^{\frac{\alpha}{2}}}+\frac{1}{2^{\frac{\alpha}{2}}}+\cdots+\frac{1}{n^{\frac{\alpha}{2}}}}{\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{\frac{1}{(n+1)^{\frac{\alpha}{2}}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}+\sqrt{n}} {(n+1)^{\frac{\alpha}{2}}}=0$$