limit of the sequence $(n+1)^\alpha - n^\alpha$

Question

Let $u_n$ be defined as : $u_n = (n+1)^{\alpha} - n^\alpha$. Now I would like to compute $\lim_{n \to \infty} u_n$ (this limit depends on the choice of $\alpha$).

Here is what I found so far :

If $\alpha \in [0,1[$ then $\lim_{n \to \infty} u_n = 0$. If $\alpha = 1$ then $\lim_{n \to \infty} u_n = 1$. If $\alpha \geq 2$, $\lim_{n \to \infty} u_n = +\infty$.

Now for $\alpha = 1$ and $\alpha \geq 2$ it's really easy to get the result. For $\alpha \in [0,1)$ I found it hard and here is what I do :

we use the mean value theorem to say that : $$\mid (n+1)^\alpha - n^\alpha \mid \leq \alpha n^{\alpha-1} \to 0$$ And we get the desired result. Now my question is :

1- Is it possible to get the limit when $\alpha\in [0,1)$ with an other technique (without using the mean value theorem) ?

2- How to get the limit when $\alpha \in (1, 2)$ ?

Thank you !

The $\alpha n^{1-\alpha}$ above should be $\alpha n^{\alpha-1}$ — robjohn, Mar 18 '19 at 12:34
@robjohn Thank you for noticing this. It shoud be correct now — mouargmouarg, Mar 18 '19 at 12:48
@dmtri Thank you for catching the mistake. And yes it mean $[0,1)$. — mouargmouarg, Mar 18 '19 at 12:49
Why not let $x=1/n$ and use L'Hopital on $(1+x)^a-1\over x^a$ as $x\to0^+$? It seems to me you get $+\infty$ for the limit as soon as $a\gt1$. — Barry Cipra, Mar 18 '19 at 12:59
Similar: https://math.stackexchange.com/questions/2276667/for-alpha-in-0-frac32-define-x-n-n1-alpha-n-alpha-then?rq=1 — StubbornAtom, Mar 18 '19 at 13:25

score 3 · Answer 1 · edited Jun 12 '20 at 10:38

3

If $\alpha\le 0$ the limit is $0$. Suppose $\alpha > 0$.

We write: $$u_n=n^{\alpha}(1-\frac{1}{n})^{\alpha}=n^{\alpha}\left(-\frac{\alpha}{n}+o(\frac{1}{n^2})\right)$$ $$=\alpha n^{\alpha-1}+o(n^{\alpha-2})$$ We conclude that:

If $0<\alpha<1, u_n\longrightarrow 0$

If $\alpha>1, u_n\longrightarrow \infty$

if $\alpha=1$ the limit is 1.

edited Jun 12 '20 at 10:38

Community

1

answered Mar 18 '19 at 14:09

DINEDINE

6,218

score 1 · Answer 2 · answered Mar 18 '19 at 13:02

1

Hint: Notice that $$\alpha n^{\alpha-1} \leq u_n\leq \alpha(n+1)^{\alpha-1}.$$ Hence by the Sandwich theorem, $$\lim \alpha n^{\alpha-1}\leq\lim u_n\leq\lim\alpha n^{\alpha-1}$$ since $\lim \alpha n^{\alpha-1}=\lim\alpha (n+1)^{\alpha-1}$. It remains to find $\lim\alpha n^{\alpha-1}$.

answered Mar 18 '19 at 13:02

YiFan Tey

17,909

How did you obtain that estimate? Note that OP asked for a technique “without using the mean value theorem.” – Martin R Mar 18 '19 at 13:39
@MartinR There's no need for MVT. Just note that the derivative of the function is monotonous (by looking at the second derivative) so that we have the desired bound by the Intermediate Value Theorem. – YiFan Tey Mar 18 '19 at 13:43

19021605 · Answer 3 · 2025-09-20T13:57:51.547

The problem, if $\alpha\in(0,1)$, is problem 2.2.3 in W. J. Kaczor and M. T. Nowak's Problems in Mathematical Analysis I (page 27). The following solution for this case can be found in p.163 of the same book.

Suppose $\alpha\in(0,1)$. We have $$0\leq(n+1)^{\alpha}-n^{\alpha}=n^{\alpha}\left(\left(1+\frac{1}{n}\right)^{\alpha}-1\right)\leq n^{\alpha}\left(1+\frac{1}{n}-1\right)=\frac{1}{n^{1-\alpha}}.$$ By Sequeeze Theorem, $\lim_{n\to\infty}(n+1)^{\alpha}-n^{\alpha}=0$.

Similarly, if $\alpha\in(1,+\infty)$, we have $$(n+1)^{\alpha}-n^{\alpha}\geq n^{\alpha}\left(\left(1+\frac{1}{n}\right)^{\alpha}-1\right)\geq n^{\alpha}\left(1+\frac{1}{n}-1\right)=n^{\alpha-1}.$$ Therefore, $\lim_{n\to\infty}(n+1)^{\alpha}-n^{\alpha}=+\infty$.

limit of the sequence $(n+1)^\alpha - n^\alpha$

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