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Let $a_n=(n+1)^c - n^c$ for some real number $0. 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 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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    @DidierPiau No, I ended up writing a similar version with little-o in the comments. Still haven't covered big-O unfortunately.2011-10-12

2 Answers 2

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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. @Did'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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    @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).2011-10-12
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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.2011-10-12