I have a series:
$$\ \sum_{n=1}| \tan (\sin(\frac{1}{\sqrt n}))-\frac{1}{\sqrt n}|^a~~ a>0$$
To estimate the convergence depending on a, ~using Taylor series. I have calculated the difference in the module using Taylor series:
$$\ \frac{1}{36n^4}(1-\frac{1}{18\sqrt n}) + \frac{1}{6n\sqrt n}(1-\frac{1}{n}) $$
What can I do else? How can I solve it? Is it wrong? How can I determine the convergence?
Since
$$ \tan \left( {\sin \left( {\sqrt x } \right)} \right) - \sqrt x = \frac{{x^{3/2} }} {6} + o\left( {x^{3/2} } \right) $$ you have that $$ a_n=\left( {\sin \left( {\sqrt {\frac{1} {n}} } \right)} \right) - \sqrt {\frac{1} {n}} = \frac{1} {{6n^{3/2} }} + o\left( {\frac{1} {{n^{3/2} }}} \right) $$ Thus you have to consider the series $$ \sum\limits_{n = 1}^\infty {\frac{1} {{n^{\left( {3/2} \right)a} }}} $$ This is convergent as log as $3a/2 >1$ while is divergent if $3a/2 \leq 1$. Thus your series is convergent as long as $a> 2/3$. your calculation are the same of mine but it missing the error term.