Limit of a sequence - Apostol 10.22 #1 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 $\frac 1 2 \le c <1$.
 A: $$
0\leqslant (n+1)^c-n^c=c\int\limits_{n}^{n+1}\frac{\mathrm dx}{x^{1-c}}\leqslant\frac{c}{n^{1-c}}.
$$
A: 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,
$$
(n+1)^c-n^c = n^c \left[ \left( 1 + \frac{1}{n} \right)^c - 1 \right] = n^c \left( \frac{c}{n} + o\left( \frac 1n \right) \right) = c n^{c-1} (1+o(1)).$$
