Determine the convergence of ∑ ( (na + 1)1/2 - (na)1/2

Prove that if $a > 2$ then the series

$\sum_{n=0}^{\infty} \left( \sqrt{n^a + 1} - \sqrt{n^a} \right)$

converges, and that it diverges if $a = 2$ .

Proof. We have

$\begin{align*} \sqrt{n^a + 1} - \sqrt{n^a} &= \frac{\left( \sqrt{n^a+1} - \sqrt{n^a} \right) \left( \sqrt{n^a + 1} + \sqrt{n^a} \right)}{\sqrt{n^a+1} + \sqrt{n^a}} \\[9pt] &= \frac{1}{\sqrt{n^a + 1} + \sqrt{n^a}}. \end{align*}$

Then, we use the limit comparison test with the series $\sum \frac{1}{n^{\frac{a}{2}}}$ .

$\begin{align*} \lim_{n \to \infty} \frac{ \frac{1}{n^{\frac{a}{2}}}}{\frac{1}{\sqrt{n^a+1} + \sqrt{n^a}}} &= \lim_{n \to \infty} \left( \sqrt{ 1 + \frac{1}{n^a}} + \sqrt{1} \right) \\[9pt] &= 2. \end{align*}$

Thus, the given series converges or diverges as $\sum \frac{1}{n^{\frac{a}{2}}}$ does. But, we know $\sum \frac{1}{n^{\frac{a}{2}}}$ converges for $a > 2$ and diverges for $a = 2$ as desired $. \qquad \blacksquare$

Stumbling Robot

Determine the convergence of ∑ ( (n^a + 1)^1/2 – (n^a)^1/2

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