Determine whether ∑ ns((n+1)1/2 - 2n1/2 + (n-1)1/2) converges

Test the following series for convergence:

$\sum_{n=1}^{\infty} n^s \left( \sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1} \right).$

The series is convergent if $s < \frac{1}{2}$ and divergent if $s \geq \frac{1}{2}$ .

Proof. We will use the limit comparison test with series $\sum \frac{1}{n^{\frac{3}{2} - s}}$ (which we know converges for $s < \frac{1}{2}$ and diverges for $s \geq \frac{1}{2}$ ). First, we rewrite the terms in the series,

$\begin{align*} \left|n^s \left( \sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1} \right) \right|&= n^s \left| \left( \frac{ \left( \sqrt{n+1} - 2\sqrt{n} + \sqrt{n-1}\right)\left(\sqrt{n+1} + 2 \sqrt{n} + \sqrt{n-1} \right)}{ \sqrt{n+1} + 2\sqrt{n} + \sqrt{n-1}} \right) \right|\\[9pt] &= n^s \left| \left( \frac{2 \sqrt{n+1}\sqrt{n-1} - 2n}{\sqrt{n+1} + 2\sqrt{n} + \sqrt{n-1}} \right) \right| \\[9pt] &= n^s \left| \left( \frac{2 \left( \sqrt{n^2 - 1} - n \right)}{2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1}} \right) \right|\\[9pt] &= n^s \left| \left( \frac{2 \left( \sqrt{n^2 - 1} - n \right) \left( \sqrt{n^2-1} + n \right)}{\left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right) \right|\\[9pt] &= n^s \left|\left( \frac{-2}{\left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right)\right| \\[9pt] &= n^2 \left( \frac{2}{ \left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right). \end{align*}$

Now, we can use the limit comparison test with $\frac{1}{n^{\frac{3}{2} - s}}$ .

$\begin{align*} & \lim_{n \to \infty} \frac{ n^s \left( \frac{2}{\left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right) }{ \frac{1}{n^{\frac{3}{2}-s}} } \\[9pt] &= \lim_{n \to \infty} n^{\frac{3}{2}} \left( \frac{2}{\left( 2 \sqrt{n} + \sqrt{n+1} + \sqrt{n-1} \right) \left( \sqrt{n^2-1} + n \right)} \right) \\[9pt] &= \lim_{n \to \infty} \frac{2}{\left( 2 + \sqrt{1 + \frac{1}{n}} + \sqrt{1 - \frac{1}{n}} \right) \left( \sqrt{1 - \frac{1}{n^2}} + 1 \right)} \\[9pt] &= \frac{1}{4}. \end{align*}$

(In the second to last line, we multiplied the numerator and denominator by $n^{-\frac{3}{2}}$ and in the denominator we had $n^{-\frac{3}{2}} = n^{-\frac{1}{2}}\cdot n^{-1}$ and split these two pieces between the two terms in the product of the denominator.) Therefore, the given series converges or diverges as $\sum \frac{1}{n^{s-\frac{3}{2}}}$ does. Since $\sum \frac{1}{n^{s - \frac{3}{2}}}$ converges for $s < \frac{1}{2}$ and diverges for $s \geq \frac{1}{2}$ , we have proved our claim $. \qquad \blacksquare$