Convergent sequence or not? Let integer $p>1$ and   $a_n=n^{-\frac{p-1}{p}}\sum_{i=1}^n(i+1)^{-\frac1p}$. Then $a_n$ is convergent sequence or not ? If $a_n$ is convergent sequence, then $\lim_{n\to \infty}a_n= \frac{p-1}{p}$ or not?
 A: 
The standard comparison with an integral does the job with little effort. 

To wit, for every $p\gt0$, the function $u:x\mapsto x^{-1/p}$ is decreasing on $x\gt0$ hence, for every $i\geqslant1$, for every $x$ in the interval $((i+1)/n,(i+2)/n)$ and every $y$ in the interval $(i/n,(i+1)/n)$, 
$$
u(x)\leqslant u((i+1)/n)\leqslant u(y).
$$
Integrating this double inequality yields
$$
n\int_{(i+1)/n}^{(i+2)/n}u(x)\mathrm dx\leqslant u((i+1)/n)\leqslant n\int_{i/n}^{(i+1)/n}u(y)\mathrm dy.
$$
On the other hand,
$$
a_n=n^{-(p-1)/p}n^{-1/p}\sum_{i=1}^nu((i+1)/n)=n^{-1}\sum_{i=1}^nu((i+1)/n),
$$
hence, summing the estimations of $u((i+1)/n)$ above yields
$$
\int_{2/n}^{(n+2)/n}u(x)\mathrm dx\leqslant a_n\leqslant \int_{1/n}^{(n+1)/n}u(x)\mathrm dx.
$$
If $p\ne1$, a primitive of $u$ is $v:x\mapsto (p/(p-1))x^{1-1/p}$. One sees that $v(1/n)$ and $v(2/n)$ both converge to $v(0)=0$ (this is the first and only time when one uses the hypothesis that $p\gt1$...) and that $v((n+1)/n)$ and $v((n+2)/n)$ both converge to $v(1)=p/(p-1)$. Finally,
$$
\lim_{n\to\infty}a_n=p/(p-1).
$$
A: One can use a Riemann sum argument. But we have to be careful as the map $x^{-1/p}$ cannot be extended by continuity on $[0,1]$. But it's not a problem, as for a fixed $a$, 
$$\frac 1n\sum_{j=\lfloor na\rfloor}^n(j/n)^{-1/p}\to \int_a^1x^{-1/p}\mathrm dt$$
and $$\frac 1n\sum_{j=1}^{\lfloor na\rfloor}(j/n)^{-1/p}=\frac 1n\sum_{j=1}^{\lfloor na\rfloor}(n/j)^{1/p}\leqslant n^{1/p-1}(1+\sum_{j=2}^{\lfloor na\rfloor}\int_{j-1}^js^{-1/p}\mathrm ds)=n^{1/p-1}\left[\frac 1{1-1/p}s^{1-1/p}\right]_1^{\lfloor na\rfloor}\leqslant C_p(a+1/n),$$ where $C_p$ depends only on $p$. Now we conclude by a $2\varepsilon$-argument: take $a$ such that $\int_0^at^{-1/p}\mathrm dt\lt\varepsilon$ and do the job for $\int_a^1t^{-1/p}\mathrm dt$.
