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Let $a_n=(n+1)^c - n^c$ for some real number $0<c<1$. Prove that this sequence converges, and calculate the limit.

I have already proved it converges by showing it is bounded and monotonically decreasing. I think that showing that the limit is 0 for $0<c<\tfrac 1 2$ is easy:

$$\lim_{n\to\infty} (n+1)^c-n^c= \lim_{n\to\infty}\frac{(n+1)^{2c}-n^{2c}}{(n+1)^c+n^c}=0$$ Since, for these values of c, $\frac d {dx}(n+1)^{2c}-n^{2c}<0$.

I would appreciate a hint for how to deal with $\frac 1 2 \le c <1$.

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25 minutes. $ $ –  Did Oct 12 '11 at 6:33
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@Didier I am a little curious. What does your comment ("25 minutes.") mean? –  Srivatsan Oct 12 '11 at 15:20
    
@Srivatsan, my comment was a concise way of mentioning that the OP accepted an answer quite shortly after having posted the question. Since this is not the first time, I thought worthwhile to (try to) draw the OP's attention to the (obvious) drawbacks of these hasty procedures which seem to run contrary to the purpose (and usual practices) of the site. –  Did Oct 12 '11 at 18:55
    
@DidierPiau Thanks for bringing this to my attention. I was under the impression that as soon as I receive an answer which I understand, I should accept it. In the future I will give it more time before accepting an answer. –  Michael Boratko Oct 12 '11 at 19:00
    
@process, an answer which I understand... So now, you do understand big-O stuff? Nice to know. –  Did Oct 12 '11 at 19:09
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2 Answers

up vote 3 down vote accepted

I cannot think of an easy way to push the OP's approach to completion. Here's a fresh approach.

Since $$ \left( 1 + \frac{1}{n} \right)^c - 1 = O\left(\frac{1}{n} \right) \tag{1} $$ for any fixed $c \in (0, 1)$, it follows that $$ (n+1)^c - n^c = O(n^{c-1}) = o(1). $$

Actually, $(1)$ needs some justification, which might be suppressed depending on the context (and the level of the author/intended audience).

A solution without big-Oh. As per the OP's comment below, here's a way to remove the use of the big-Oh notation in the proof. Didier's answer gives a similar proof that in fact gives a tighter estimate.

If $0< c < 1$, then we have: $$ (n+1)^c-n^c = n^c \left[ \left( 1 + \frac{1}{n} \right)^c - 1 \right] \leq n^c \left[\left( 1 + \frac{1}{n} \right)^{1} - 1 \right] = n^{c-1}. $$ We can proceed exactly as before.

Using Taylor expansion. This solution is from the OP's comment below. For small $x$, by Taylor expansion, we have $(1+x)^c = 1+cx + o(x)$. Therefore, $$ (n+1)^c-n^c = n^c \left[ \left( 1 + \frac{1}{n} \right)^c - 1 \right] = n^c \left( \frac{c}{n} + o\left( \frac 1n \right) \right) = c n^{c-1} (1+o(1)).$$

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I wish I could... it seems like all the problems I get stuck on have a simple solution with big-O notation, but the book has not covered this so I don't want to go "out of sequence" :) –  Michael Boratko Oct 12 '11 at 5:52
    
We have covered little-o notation, how about this: $(1+x)^c-1=cx+o(x)$ (Taylor series for $x\to0$) and so $(n+1)^c-n^c=cn^{c-1}+o(n^{c-1})$. This is basically equivalent to your solution, I think. I wish Apostol covered big-O notation though... –  Michael Boratko Oct 12 '11 at 5:59
    
@process Your solution is correct as well. In fact, it is a slightly more refined estimate as compared to mine (both of them work for this question). –  Srivatsan Oct 12 '11 at 6:02
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$$ 0\leqslant (n+1)^c-n^c=c\int\limits_{n}^{n+1}\frac{\mathrm dx}{x^{1-c}}\leqslant\frac{c}{n^{1-c}}. $$

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This is a nice solution too. –  Michael Boratko Oct 12 '11 at 6:41
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