Which exponents r>0 is the limit finite I am trying to find values of $r>0$ such that $\lim\limits_{n\rightarrow \infty} \sum\limits_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}$ is finite. I have tried to use integral methods for this limit such as $\sum\limits_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}=\sum\limits_{k=1}^{n}\frac{n^{r-1}}{n^r+k^r}+\sum\limits_{k=n}^{n^2}\frac{n^{r-1}}{n^r+k^r} = \frac1n\sum\limits_{k=1}^{n}\frac{1}{1+(\frac kn)^r}+\sum\limits_{k=n}^{n^2}\frac{n^{r-1}}{n^r+k^r}$. So the first term, I could use the integral method to estimate. However, the second one is hard to evaluate. Could you please help me with this? Thanks

In short: diverges for $r\in [0,1]$, converges for $r>1$.


*

*First case (simpler): $r\in(0,1)$:


$$
\sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r} = \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^r+k^r}
\geq \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^r+n^{2r}}
\geq \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^{2r}+n^{2r}}
= \frac{n^2}{n}\frac{1}{2n^r} = \frac{n^{1-r}}{2}
$$
so for $r\in[0,1)$ the sequence will diverge by comparison.


*

*Second case: for $r=1$,
$$\begin{align}
\sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r} &= 
\sum_{k=1}^{n^2}\frac{1}{n + k} = 
\sum_{k=1}^{n}\frac{1}{n + k} +  \sum_{k=n+1}^{n^2}\frac{1}{n + k}\\
&= \underbrace{\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1 + \frac{k}{n}}}_{\text{Converges (Riemann sum)}} +  \sum_{k=n+1}^{n^2}\frac{1}{n + k}
\end{align}$$
but
$$
\sum_{k=n+1}^{n^2}\frac{1}{n + k} = 
\sum_{\ell=1}^{n-1} \sum_{k=\ell n+1}^{(\ell+1)n}\frac{1}{n + k}
\geq \sum_{\ell=1}^{n-1}  \sum_{k=\ell n+1}^{(\ell+1)n}\frac{1}{(\ell+2)n}
= \sum_{\ell=1}^{n-1} \frac{1}{\ell+2} \sim_{n\to\infty} \ln n
$$
so 
$$
\sum_{k=1}^{n^2}\frac{1}{n+k} \xrightarrow[n\to\infty]{} \infty
$$

*Last case: $r>1$. Same technique, and as before we only need to analyze the second term.
$$\begin{align}
\sum_{k=n+1}^{n^2}\frac{n^{r-1}}{n^r+k^r}
&= \frac{1}{n}\sum_{k=n+1}^{n^2}\frac{1}{1+\left(\frac{k}{n}\right)^r}
= \frac{1}{n}\sum_{\ell=1}^{n-1} \sum_{k=\ell n+1}^{(\ell+1)n} \frac{1}{1+\left(\frac{k}{n}\right)^r}
\leq
\frac{1}{n}\sum_{\ell=1}^{n-1} \sum_{k=\ell n+1}^{(\ell+1)n} \frac{1}{1+\ell^r}
\\
&=
\sum_{\ell=1}^{n-1} \frac{1}{1+\ell^r}
\end{align}$$
which therefore converges by comparison ($p$-series with exponent $r>1$). 
$$
\exists L_r <\infty,\qquad \sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r} \xrightarrow[n\to\infty]{} L_r
$$
