Test the series: infinity sigma n=1 (-1)^ n-1 sin nx n sqrt n for absolute and conditional convergence
"\\sum (-1)^ {n-1}\\frac{ sin nx}{ n \\sqrt {n}}"
"a_n=|(-1)^ {n-1}\\frac{ sin nx}{ n \\sqrt {n}}|=\\frac{ |sin nx|}{ n \\sqrt {n}}"
"b_n=\\frac{1}{n\\sqrt n}"
since "a_n\\le b_n" and "\\sum b_n" converges , "\\sum a_n" converges as well
so, series "\\sum (-1)^ {n-1}\\frac{ sin nx}{ n \\sqrt {n}}" converges absolutely
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