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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

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