Convergent series after rearrangement becomes divergent Show that $$1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-\cdots$$
is convergent but by rearrangement the following series
 $$\left(1+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{7}}-\frac{1}{\sqrt{4}}\right)+\cdots $$
 is divergent.
Attempt:

The 1st series $$\sum_{n=1}^\infty (-1)^{n+1} \frac{1}{\sqrt{n}}$$ can be proved to be convergent by Leibnitz Test as
(1) $u_n= \frac{1}{\sqrt{n}}\to 0$ as $n\to\infty$ 
(2) $\{u_n\}$ is monotone decreasing

But please help me to show that the second series, after rearrangement, is divergent.
 A: Hint: for $n\ge 0:$
$$\dfrac{1}{\sqrt{4n+1}} + \dfrac{1}{\sqrt{4n+3}} - \dfrac{1}{\sqrt{2(n+1)}}>\dfrac{1}{\sqrt{4n+4}} + \dfrac{1}{\sqrt{4n+4}} - \dfrac{1}{\sqrt{2(n+1)}}=\\
\left(1-\frac{1}{\sqrt{2}}\right)\cdot \frac{1}{\sqrt{n+1}}$$
A: Your question is: prove that the series$$\sum_{n=1}^\infty\frac1{\sqrt{4n-1}}+\frac1{\sqrt{4n-3}}-\frac1{\sqrt{2n}}$$diverges. This is true because$$\lim_{n\to\infty}\frac{\frac1{\sqrt{4n-1}}+\frac1{\sqrt{4n-3}}-\frac1{\sqrt{2n}}}{\frac1{\sqrt{n}}}=\frac12+\frac12-\frac1{\sqrt2}=1-\frac1{\sqrt2}$$and the series $\sum_{n=1}^\infty\frac1{\sqrt{n}}$ diverges.
A: $$\sum_{n\ge 0} \left( \dfrac{1}{\sqrt{4n+1}} + \dfrac{1}{\sqrt{4n+3}} - \dfrac{1}{\sqrt{2(n+1)}} \right)$$
Find a common denominator:
$$\dfrac{\sqrt{(4n+3)(2n+2)}+\sqrt{(4n+1)(2n+2)}-\sqrt{(4n+1)(4n+3)}}{\sqrt{(4n+1)(4n+3)(2n+2)}}$$
Show that this is always positive (so it is not an alternating series), then use the comparison test with a divergent series to show that the series diverges. Hint: show that the fraction above is always greater than
$$\dfrac{1}{k\sqrt{n}}$$
For some arbitrarily large constant $k$.
