I've been studying Buck's Advanced Calculus and I have stumbled upon this question in the chapter "Conditionally Convergent Series". I need to investigate the convergence of the series given below:
$$1+ \frac1{\sqrt3} - \frac1{\sqrt2} + \frac1{\sqrt5} + \frac1{\sqrt7} - \frac1{\sqrt4} + \frac1{\sqrt9} + \frac1{\sqrt{11}} - \frac1{\sqrt6} + \cdots$$
Hint: Use Dirichlet test.
I have tried using summation by parts method, but i couldn't get it done. Then i've tried to find a series to use in comparison test but failed miserably. Any hint, solution will be appreciated. Thanks in advance.
You can consider the $-1/\sqrt{2}$, $-1/\sqrt{4}$, $-1/\sqrt{6}$, ... as $-\sqrt{2}/\sqrt{4}$, $-\sqrt{2}/\sqrt{8}$, $-\sqrt{2}/\sqrt{12}$. That is, $$\sum_{n=1}^\infty \frac{a_n}{\sqrt{n}}$$ where $$ a_n = \cases{1 & if $n$ is odd\cr -\sqrt{2}& if $n$ is divisible by $4$\cr 0 & otherwise}$$ Now in this case $\sum_n a_n = +\infty$, so Dirichlet's test does not apply. Instead,
$$ \frac{1}{\sqrt{4n-1}} - \frac{\sqrt{2}}{\sqrt{4n}} + \frac{1}{\sqrt{4n+1}} \ge \frac{2-\sqrt{2}}{\sqrt{4n}}$$
and the comparison test show that this diverges.