For $\alpha \in (0,\frac{3}{2})$, define $x_n=(n+1)^{\alpha}-n^{\alpha}$. Then $\lim_{n \to \infty}x_n$ is?

Question

For $\alpha \in (0,\frac{3}{2})$, define $x_n=(n+1)^{\alpha}-n^{\alpha}$. Then $\lim_{n \to \infty}x_n$ is?

A) 1 for all $\alpha$

B) 1 or 0 depending on the value of $\alpha$

C) 1 or $\infty$ depending on the value of $\alpha$

D) 1 , 0 or $\infty$ depending on the value of $\alpha$

Now here I am able to get 1 and $\infty$ depending on the value of $\alpha$ but how does 0 counts in? Since it's possible to get 0 only when value of $\alpha$ is 0 but here here 0 is not included in interval.

What am I doing wrong?

Answer 1

We can evaluate the limit a number of ways. Herein, we present three approaches.

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METHODOLOGY $1$: Apply Standard Inequalities and the Squeeze Theorem

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function and the logarithm function satisfy the inequalities

$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$

for $x<1$ and

$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 2$$

for $x>0$.

First we write

$$\begin{align} (n+1)^\alpha-n^\alpha&=e^{\alpha\log(n)}\left(e^{\alpha \log\left(1+\frac1n\right)}-1\right)\tag3 \end{align}$$

Next, we use $(1)$ and $(2)$ in $(3)$ to obtain

$$\frac{\alpha n^{\alpha-1}}{1+1/n}\le (n+1)^\alpha-n^\alpha\le \frac{\alpha n^{\alpha-1}}{1-\alpha/n}\tag 4$$

whereupon applying the squeeze theorem to $(4)$ reveals

$$\begin{align} \lim_{n\to \infty }\left((n+1)^\alpha-n^\alpha\right)&= \begin{cases}\infty&,1<\alpha<3/2\\\\ 1&,1=\alpha\\\\ 0&\alpha<1 \end{cases} \end{align}$$

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METHODOLOGY $2$: Asymptotic Analysis

Here, we expand $(n+1)^\alpha$ using the general binomial theorem to find

$$\begin{align} (n+1)^\alpha-n^\alpha&=n^\alpha\left(\left(1+\frac1n\right)^\alpha-1\right)\\\\ &=n^\alpha\left(\left(1+\frac\alpha n+O\left(\frac1{n^2}\right)\right)-1\right)\\\\ &=\alpha n^{\alpha -1}\\\\ &\to \begin{cases}\infty&,1<\alpha<3/2\\\\ 1&,1=\alpha\\\\ 0&\alpha<1 \end{cases} \end{align}$$

as expected!

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METHODOLOGY $3$: Apply the Mean Value Theorem

Finally, applying the mean value theorem to the function $f(x)=x^\alpha$ yields

$$(x+1)^\alpha-x^\alpha=\alpha \xi^{\alpha-1}$$

for $x<\xi<x+1$. Letting $x\to \infty$, we find again that

$$\begin{align} \lim_{x\to \infty}\left((x+1)^\alpha-x^\alpha\right)&= \begin{cases}\infty&,1<\alpha<3/2\\\\ 1&,1=\alpha\\\\ 0&\alpha<1 \end{cases} \end{align}$$

as expected!

Answer 2

$$(n+1)^{\frac{1}{2}}-n^{\frac{1}{2}}=\frac{n+1-n}{(n+1)^{\frac{1}{2}}+n^{\frac{1}{2}}}=\frac{1}{(n+1)^{\frac{1}{2}}+n^{\frac{1}{2}}}\to0 \quad \text{as }n\to\infty$$

Answer 3

If you know the mean value theorem, then let $f(x)=x^{\alpha}.$

By the mean value theorem, $f(n+1)-f(n)=f'(c_n)$ for some $c_n\in(n,n+1)$.

But $f'(x)=\alpha x^{\alpha-1}$. So we have that:

$$f(n+1)-f(n)=\alpha c_n^{\alpha-1}, c_n\in(n,n+1)$$

Consider the three cases: $\alpha<1,\alpha=1,\alpha>1$.

I'll give you a start.

If $\alpha>1$ then $f'(x)$ is increasing, and $f(n+1)-f(n)=f'(c_n)> f'(n)=\alpha n^{\alpha-1}$. Use this to prove $f(n+1)-f(n)\to \infty$.