Compute limits of (n+1)c - nc for real values of c

Compute the limit
$\lim_{n \to \infty} a_n \qquad \text{where} \qquad a_n = \sqrt{n+1} - \sqrt{n}.$
Compute the limit
$\lim_{n \to \infty} a_n \qquad \text{where} \qquad a_n = (n+1)^c - n^c$

for $c \in \mathbb{R}$ . Determine the values for which the sequences diverges and for which it converges and compute the values of the limits in the convergent case.

We multiply and divide the terms by $\sqrt{n+1} + \sqrt{n}$ to get,
$\begin{align*} \lim_{n \to \infty} \left( \sqrt{n+1} - \sqrt{n} \right) &= \lim_{n \to \infty} \left( \frac{\left( \sqrt{n+1} - \sqrt{n} \right)\left( \sqrt{n+1} + \sqrt{n} \right)}{\sqrt{n+1} + \sqrt{n}}\right) \\[9pt] &= \lim_{n \to \infty} \frac{1}{\sqrt{n+1} + \sqrt{n}} \\[9pt] &= 0. \end{align*}$
Observe that
$(n+1)^c - n^c = c \int_n^{n+1} \frac{1}{x^{1-c}} \, dx$

since

$c \int_n^{n+1} \frac{1}{x^{1-c}} \,dx = c \cdot \left( \frac{x^c}{c} \Bigr \rvert_n^{n+1} \right) = (n+1)^c - n^c.$

But then, if $0 < c < 1$ , the function inside the integral is a decreasing function; hence,

$c \int_n^{n+1} \frac{1}{x^{1-c}} \, dx \leq \frac{c}{n^{1-c}}.$

Since this limit goes to 0 for $0 < c < 1$ (since $1-c > 0$ ) we have that the sequence converges to 0 for these values of $c$ .
If $c > 1$ then the integrand is increasing, and so,

$c \int_n^{n+1} \frac{1}{x^{1-c}} \, dx \geq \frac{c}{n^{1-c}}.$

In this case the limit of $\frac{c}{n^{1-c}}$ as $n \to \infty$ diverges to $+\infty$ (since $1-c < 0$ ). Hence, the sequence does not converge.
Finally, if $c= 1$ then

$\lim_{n \to \infty} ((n+1)^c - n^c) = \lim_{n \to \infty} 1 = 1.$

Putting this all together we have

$\begin{align*} \{ a_n \} &\text{ converges to } 0 \text{ for } 0 < c < 1 \\[9pt] \{ a_n \} & \text{ converges to } 1 \text{ for } c = 1 \\[9pt] \{ a_n \} & \text{ diverges for } c > 1. \end{align*}$