Consider two series \(\mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1\;}}{a_n}\) and \(\mathop \sum \limits_{n = 2}^\infty {\left( { - 1} \right)^{n - 1\;}}{b_n}\), where \({a_n} = \frac{1}{{\sqrt n }},\;{b_n} = \frac{{{x^n}}}{{n\left( {n - 1} \right)}}\) 0 < x < 1. Then:

• A. \(\mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1\;}}{a_n}\) is convergent but \(\mathop \sum \limits_{n = 2}^\infty {\left( { - 1} \right)^{n - 1\;}}{b_n}\) is divergent.
• B. both series are convergent
• C. ​​​\(\mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1\;}}{a_n}\) is divergent but \(\mathop \sum \limits_{n = 2}^\infty {\left( { - 1} \right)^{n - 1\;}}{b_n}\) is convergent
• D. both series are divergent.