Find the value of $\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{1}{n+k^{\alpha}}\right)$ 
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.
 A: 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
$$\lim_{n\rightarrow\infty}\frac{\frac{1}{1^{\frac{\alpha}{2}}}+\frac{1}{2^{\frac{\alpha}{2}}}+\cdots+\frac{1}{n^{\frac{\alpha}{2}}}}{\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{\frac{1}{(n+1)^{\frac{\alpha}{2}}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}+\sqrt{n}}  {(n+1)^{\frac{\alpha}{2}}}=0$$
A: First suppose $a\in (0,1]$. Then
$${n\over {n+1}}\le \sum_{k=1}^n {1\over {a^k+n}}\le 1$$
and it follows that the limit equals $1$.
Now suppose $a>1$. Fixing $\epsilon>0$, we have
$$\sum_1^n {1\over {n+a^k}}=\sum_1^{\epsilon n} {1\over {n+a^k}}+\sum_{\epsilon n}^n  {1\over {n+a^k}}\le \epsilon+{n\over {a^{\epsilon n}}}<2\epsilon$$
when $n$ is large. Hence the limit equals $0$.
