Find this limit $\lim_{n\to\infty}\frac{1}{n^{1+\alpha}}(a_{1}+a_{2}+\cdots+a_{n})$ let sequence $\{a_{n}\}$ such
$$\lim_{n\to\infty}\dfrac{a_{n}}{n^{\alpha}}=1(\alpha>0)$$
Useing Riemann integral of suitably chosen  functions,Find  the following limit
$$I=\lim_{n\to\infty}\dfrac{1}{n^{1+\alpha}}(a_{1}+a_{2}+\cdots+a_{n})$$
If  this problem can use Stloz lemma: we have
$$I=\lim_{n\to\infty}\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n^{1+\alpha}}=\lim_{n\to\infty}\dfrac{a_{n}}{n^{1+\alpha}-(n-1)^{1+\alpha}}=\dfrac{1}{\alpha+1}$$
becasuse
$$\lim_{n\to\infty}\dfrac{a_{n}}{n^{\alpha}}=1(\alpha>0)$$
But use Riemann integral of suitably chosen  functions: I have
$$I=\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}\dfrac{a_{i}}{n^a}$$
I guess we will prove
$$I=\int_{0}^{1}x^{\alpha}dx=\dfrac{1}{1+\alpha}$$
But I can't prove this equation.
Thank you
 A: Define
$$
S_n \equiv \frac{1}{n}\sum_{i=1}^n \frac{a_i}{n^\alpha}.
$$
Fix $\epsilon>0$ and assume $\epsilon<1$. Since $\lim_{n\to\infty} \frac{a_n}{n^\alpha} = 1$, there is an $N$ such that $n>N$ implies
$$
1 - \epsilon < \frac{a_n}{n^\alpha} < 1 + \epsilon.
$$
Since
$$
\frac{a_i}{n^\alpha} = \frac{a_i}{i^\alpha}\cdot\frac{i^\alpha}{n^\alpha},
$$
for $n>N$,
$$
(*) \ \ \frac{1}{n}\sum_{i=1}^N \frac{a_i-(1-\epsilon)i^\alpha}{n^\alpha} + \frac{1-\epsilon}{n}\sum_{i=1}^n \frac{i^\alpha}{n^\alpha}
< S_n < \frac{1}{n}\sum_{i=1}^N \frac{a_i-(1+\epsilon)i^\alpha}{n^\alpha} + \frac{1+\epsilon}{n}\sum_{i=1}^n \frac{i^\alpha}{n^\alpha}.
$$
Observe that
$$
\lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n \frac{i^\alpha}{n^\alpha} = \int_0^1 x^\alpha \ dx  = \frac{1}{1+\alpha}.
$$
This follows because the sum on the lhs is just the Riemann sum approximating the integral on the rhs. Now just take the limit as $n\to\infty$ of the inequalities $(*)$ to obtain
$$
\frac{1-\epsilon}{1+\alpha} \le \liminf_{n\to\infty} S_n \le \limsup_{n\to\infty} S_n \le \frac{1+\epsilon}{1+\alpha}.
$$
Since $\epsilon\in(0,1)$ was arbitrary, it follows
$$
\lim_{n\to\infty} S_n = \frac{1}{1+\alpha}.
$$
