Indefinite Integration 4 Question 4
4. For $a \in R$ (the set of all real numbers), $a \neq-1$, $\lim _{n \rightarrow \infty} \frac{\left(1^{a}+2^{a}+\ldots+n^{a}\right)}{(n+1)^{a-1}[(n a+1)+(n a+2)+\ldots+(n a+n)]}=\frac{1}{60}$. Then, $a$ is equal to
(a) 5
(b) 7
(c) $\frac{-15}{2}$
(d) $\frac{-17}{2}$
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Answer:
Correct Answer: 4. (b)
Solution:
- PLAN Converting Infinite series into definite Integral
$$ \begin{array}{ll} \text { i.e. } & \lim _{n \rightarrow \infty} \frac{h(n)}{n} \\ \begin{array}{ll} \lim _{n \rightarrow \infty} \frac{1}{n} \sum _{r=g(n)}^{h(n)} f & \frac{r}{n}=\int f(x) d x \\ & \lim _{n \rightarrow \infty} \frac{g(n)}{n} \end{array} \end{array} $$
where, $\frac{r}{n}$ is replaced with $x$.
$\Sigma$ is replaced with integral.
$$ \begin{aligned} & \lim _{n \rightarrow \infty} \frac{1^{a}+2^{a}+\ldots+n^{a}}{(n+1)^{a-1}{(n a+1)+(n a+2)+\ldots+(n a+n)}}=\frac{1}{60} \\ & \Rightarrow \lim _{n \rightarrow \infty} \frac{\sum _{r=1}^{n} r^{a}}{(n+1)^{a-1} \cdot n^{2} a+\frac{n(n+1)}{2}}=\frac{1}{60} \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \lim _{n \rightarrow \infty} \frac{2 \sum _{r=1}^{n} \frac{r}{n}}{1+\frac{1}{n}^{a-1} \cdot(2 n a+n+1)}=\frac{1}{60} \\ & \Rightarrow \lim _{n \rightarrow \infty} \frac{1}{n} 2 \sum _{r=1}^{n} \frac{r}{n}^{a} \cdot \lim _{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}^{a-1} \cdot 2 a+1+\frac{1}{n}}=\frac{1}{60} \\ & \Rightarrow 2 \int _0^{1}\left(x^{a}\right) d x \cdot \frac{1}{1 \cdot(2 a+1)}=\frac{1}{60} \\ & \Rightarrow \quad \frac{2 \cdot\left[x^{a+1}\right] _0^{1}}{(2 a+1) \cdot(a+1)}=\frac{1}{60} \\ & \therefore \quad \frac{2}{(2 a+1)(a+1)}=\frac{1}{60} \Rightarrow(2 a+1)(a+1)=120 \\ & \Rightarrow \quad 2 a^{2}+3 a+1-120=0 \Rightarrow 2 a^{2}+3 a-119=0 \end{aligned} $$
