Find the value of $a$ by evaluating the limit $L=\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\cdots+(na+n)]}=\frac{1}{60}$

Question

For $a\in\mathbb R-\{-1\}$ $$L=\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\cdots+(na+n)]}=\frac{1}{60}$$ find the value of $a$.

My attempt:

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

which as $\lim_{n\to\infty}$ must equal to $\frac{1}{a+a+a+\cdots}$. Now this obviously can't be implying some logical error on my part. What to do now?

Answer 1

We have

\begin{align} \frac1{60} &=\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\cdots+(na+n)]}\\ &= \lim_{n\to\infty}\frac{\frac1n\sum_{i=1}^n \left(\frac{i}{n}\right)^a}{\frac1n \left(\frac{n+1}n\right)^{a-1}\left(na + \sum_{i=1}^n \frac{i}{n}\right)}\\ &= \lim_{n\to\infty}\frac{\frac1n\sum_{i=1}^n \left(\frac{i}{n}\right)^a}{\left(\frac{n+1}n\right)^{a-1}\left(a + \frac{1}{n^2}\sum_{i=1}^n i\right)}\\ &= \lim_{n\to\infty}\frac{\frac1n\sum_{i=1}^n \left(\frac{i}{n}\right)^a}{\left(\frac{n+1}n\right)^{a-1}\left(a + \frac{n(n+1)}{2n^2}\right)}\\ &= \frac{\int_0^1 x^a\,dx}{1\cdot\left(a+\frac12\right)}\\ &= \frac{1}{(a+1)\left(a+\frac12\right)} \end{align}

so $a \in \left\{-\frac{17}2, 7\right\}$.

Answer 2

$L=\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+...+(na+n)]}=$

$= \lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{\frac{1}{n^a}(\frac{n+1}{n})^{a-1}\frac{1}{n}[n^2a +\frac{n(n+1)}{2}]} =$

$= \lim_{n\to\infty}\frac{(\frac{1}{n})^a+(\frac{2}{n})^a+...+1^a}{(\frac{n+1}{n})^{a-1}[na +\frac{(n+1)}{2}]} =$

$= \lim_{n\to\infty}\frac{\frac{1}{n}[(\frac{1}{n})^a+(\frac{2}{n})^a+...+1^a]}{(\frac{n+1}{n})^{a-1}[a +\frac{1}{2}+\frac{1}{n}]} =$

$= \lim_{n\to\infty}\frac{\frac{1}{n}\sum_{k=1}^{n}(\frac{k}{n})^a}{(\frac{n+1}{n})^{a-1}[a +\frac{1}{2}+\frac{1}{n}]} =$

$=\frac{\int_0^1 x^a dx}{a+\frac{1}{2}}=\frac{1}{(a+1)(a+\frac{1}{2})}$

because

$1+2+...+n=\frac{n(n+1)}{2}$