Evaluating the limit of the sequence: $\frac{ 1^a + 2^a +..... n^a}{(n+1)^{a-1}[n^2a + n(n+1)/2]}$

Question

My friend gave me this question to solve a few days ago and after I got no way to solve this, I thought I should seek some help.

I had to evaluate the limit of the following when $n$ tends to infinite.

$$ \frac{ 1^a + 2^a +..... n^a}{(n+1)^{a-1}[n^2a + n(n+1)/2]}$$

I tried to convert this limit into a definite integral but I couldn't get the expression solely in the terms of $(r/n)$

Any help on how to proceed will be appreciated.

Note: I know to solve limit of certain series by turning it into an integral from the following form: $$\sum\frac{f(r/n)}{n}$$ It turns into an integral of $f(r/n)$.

Answer 1

The given expression can be written as \begin{align*} \lim_{n\rightarrow \infty}\frac{1}{\left(1+\frac{1}{n}\right)^{a-1}}&\frac{1}{\left(a+\frac{n+1}{2n}\right)} \frac{1}{n}\sum_{r=1}^n \left(\frac{r}{n}\right)^a \\ &= \frac{1}{a+\frac{1}{2}} \int_0^1 x^a dx \\ &= \frac{2}{1+2a}\frac{1}{a+1} \\ &= \frac{2}{(a+1)(2a+1)} \end{align*}

Answer 2

Hint. One may use that, for a fixed real number $a>0$, as $n \to \infty$ we have $$ 1^a + 2^a +..... n^a \sim \frac{n^{a+1}}{a+1} \tag1 $$ giving, as $n \to \infty$, $$ \begin{align} \frac{ 1^a + 2^a +..... n^a}{(n+1)^{a-1}[n^2a + n(n+1)/2]}&\sim\frac{\frac{n^{a+1}}{a+1}}{n^2\cdot (n+1)^{a-1}\left(a + \frac12+\frac1{2n}\right)} \\\\&\sim \frac{\frac{n^{a+1}}{a+1}}{n^2\cdot n^{a-1}\left(a + \frac12\right)} \\\\&\sim \frac{1}{(a+1)\left(a + \frac12\right)} \tag2 \end{align} $$ which is the desired limit.