ISI 2016 B.Math Entrance(1)

Calculus Level 3

For α ∈ ( 0 , 3 2 ) \alpha \in \left(0,\dfrac{3}{2}\right) , define x n = ( n + 1 ) α − n α x_n = (n+1)^{\alpha}-n^{\alpha} .

What are all the possible values of lim ⁡ n → ∞ x n \displaystyle{\lim_{n \rightarrow \infty} x_n} ?

Note: If you think that the limit can also diverge to infinity , then choose the option with ∞ \infty as well.

0 , 1 , ∞ 0,1, \infty 1 , ∞ 1, \infty 0 , 1 0,1 0 , ∞ 0,\infty

Md Zuhair by Md Zuhair , 20, India

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1 solution

Tapas Mazumdar
Apr 14, 2017

We have

x n = ( n + 1 ) α − n α = ( n α + α n α − 1 + α ( α − 1 ) 2 n α − 2 + α ( α − 1 ) ( α − 2 ) 6 n α − 3 + ⋯ ) − n α = α n α − 1 + α ( α − 1 ) 2 n α − 2 + α ( α − 1 ) ( α − 2 ) 6 n α − 3 + ⋯ Using generalized binomial theorem = α n α − 1 ( 1 + ( α − 1 ) 2 n + ( α − 1 ) ( α − 2 ) 6 n 2 + ⋯ ) ⏟ B \begin{aligned} x_n = (n+1)^{\alpha} - n^{\alpha} &= \left( n^{\alpha} + \alpha n^{\alpha - 1} + \dfrac{\alpha(\alpha - 1)}{2} n^{\alpha -2} + \dfrac{\alpha(\alpha - 1)(\alpha -2)}{6} n^{\alpha -3} + \cdots \right) - n^{\alpha} \\ &= \alpha n^{\alpha - 1} + \dfrac{\alpha(\alpha - 1)}{2} n^{\alpha -2} + \dfrac{\alpha(\alpha - 1)(\alpha -2)}{6} n^{\alpha -3} + \cdots \qquad \qquad \small \color{#3D99F6}{\text{Using generalized binomial theorem}} \\ &= \alpha n^{\alpha -1} \underbrace{\left( 1 + \dfrac{(\alpha -1)}{2n} + \dfrac{(\alpha - 1)(\alpha -2)}{6n^2} + \cdots \right)}_{\large \mathfrak{B}} \end{aligned}

Note that as n → ∞ n \to \infty , B → 1 \mathfrak{B} \to 1 , so we have

lim ⁡ n → ∞ x n = lim ⁡ n → ∞ α n α − 1 \displaystyle \lim_{n \to \infty} x_n = \lim_{n \to \infty} \alpha n^{\alpha -1}

Hence

lim ⁡ n → ∞ α n α − 1 = { 0 α ∈ ( 0 , 1 ) 1 α = 1 ∞ α ∈ ( 1 , ∞ ) \displaystyle \lim_{n \to \infty} \alpha n^{\alpha -1} = \begin{cases} 0 & \alpha \in \left( 0,1 \right) \\ 1 & \alpha = 1 \\ \infty & \alpha \in \left( 1 , \infty \right) \end{cases}

The case α ∈ ( 1 , 3 2 ) \alpha \in \left(1 , \dfrac 32 \right) is just a trivial interval for the last case.

Thus the answer is 0 , 1 , ∞ \boxed{0, 1 , \infty} depending on the value of α \alpha .

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