For a (the set of all real numbers), a - 1 , lim n ( 1 a + 2 a + ? + n a ) ( n + 1 ) a - 1 [ ( n a + 1 ) + ( n a + 2 ) + ? + ( n a + n ) ] = 1 60 . Then a = [2013]

Question

For $a \in ℝ$ (the set of all real numbers), $a \neq - 1$ , $\lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{{(n + 1)}^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60} .$

Then $a =$ [2013]

Smart Achievers · Accepted Answer

(2, 4)

$Given:$

$a \neq - 1$

$\Rightarrow \lim_{n \to \infty} \frac{n^{a} [{(\frac{1}{n})}^{a} + {(\frac{2}{n})}^{a} + \dots + {(\frac{n}{n})}^{a}]}{{(n + 1)}^{a - 1} [n^{2} a + \frac{n (n + 1)}{2}]} = \frac{1}{60}$

$\Rightarrow \lim_{n \to \infty} \frac{n^{a - 1}}{{(n + 1)}^{a - 1}} \frac{\frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{a}}{[a + \frac{1}{2} (1 + \frac{1}{n})]} = \frac{1}{60}$

$\Rightarrow \lim_{n \to \infty} {(\frac{1}{1 + \frac{1}{n}})}^{a - 1} \frac{\frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{a}}{a + \frac{1}{2} (1 + \frac{1}{n})} = \frac{1}{60}$

$∵$ $\frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{a} = \int_{0}^{1} x^{a} d x$ as $\frac{1}{n} = d x$ and $\frac{r}{n} = x$

$when r = 1, n \to \infty \Rightarrow x \to 0$

$when r = n \Rightarrow x \to 1$

$\Rightarrow \frac{\int_{0}^{1} x^{a} d x}{a + \frac{1}{2}} = \frac{1}{60}$ $\Rightarrow \frac{{[x^{a + 1}]}_{0}^{1}}{(a + 1) (a + \frac{1}{2})} = \frac{1}{60}$

$\Rightarrow \frac{1}{(a + 1) (a + \frac{1}{2})} = \frac{1}{60}$

$\Rightarrow 2 a^{2} + 3 a - 119 = 0$ $\Rightarrow (a - 7) (2 a - 17) = 0$

$∴$ $a = 7 or - \frac{17}{2}$

Solution

Q.
For $a \in ℝ$ (the set of all real numbers), $a \neq - 1$ , $\lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{{(n + 1)}^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60} .$

Then $a =$ [2013]

1 $5$

2 $7$

3 $\frac{- 15}{2}$

4 $\frac{- 17}{2}$

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