If _n 1^a+2^a++n^a(n+1)^a-1[(na+2)++(na+n)] = 160 for some positive real number a, then a …

Question

If $\displaystyle\lim_{n	o\infty} \dfrac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}[(na+2)+\cdots+(na+n)]} = \dfrac{1}{60}$ for some positive real number $a$, then $a$ is equal to:

Accepted Answer

$\lim_{n	o\infty} \dfrac{1}{n}\sum_{r=1}^n \left(\dfrac{r}{n}ight)^a \left(1+\dfrac{1}{n}ight)^{a-1}\bigg/\left(a+\dfrac12\left(1+\dfrac1night)ight) = \dfrac{1}{60}$

$\Rightarrow \dfrac{\int_0^1 x^a\,dx}{a+\dfrac12} = \dfrac{1}{60} \Rightarrow \dfrac{1}{(a+1)\left(a+\dfrac12ight)} = \dfrac{1}{60}$

$\Rightarrow (a+1)(2a+1) = 120 \Rightarrow 2a^2+3a-119 = 0$

$\Rightarrow 2a^2+17a-14a-119=0 \Rightarrow (a-7)(2a+17)=0$

$\Rightarrow a = 7, -\dfrac{17}{2}$

Since $a$ is positive, $a = 7$

Solutions · 1