Question

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:
1. \(\mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1\;}}{a_n}\) is convergent but \(\mathop \sum \limits_{n = 2}^\infty {\left( { - 1} \right)^{n - 1\;}}{b_n}\) is divergent.
2. both series are convergent
3. ​​​\(\mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1\;}}{a_n}\) is divergent but \(\mathop \sum \limits_{n = 2}^\infty {\left( { - 1} \right)^{n - 1\;}}{b_n}\) is convergent 
4. both series are divergent.

Answer 1

Correct Answer - Option 2 : both series are convergent

Series \({S_1} = \mathop \sum \nolimits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{\sqrt n }}\) 

Now, \(\frac{{{t_n}}}{{{t_{n + 1}}}} = \frac{{\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{\sqrt n }}}}{{\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt {n + 1} }}}}\) 

\( = \left( { - 1} \right)\frac{{\sqrt {n + 1} }}{{\sqrt n }}\) 

\(= \left( { - 1} \right)\sqrt {1 + \frac{1}{n}} < 1\) 

And, \(\mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)\sqrt {1 + \frac{1}{n}} = - 1,\) 

Hence, the series S₁ converges.

Series \({S_2} = \mathop \sum \nolimits_{n = 2}^\infty {\left( { - 1} \right)^{n - 1}} \cdot \frac{{{{\left( x \right)}^n}}}{{n\left( {n - 1} \right)}}\) 

For 0 < x

Then, \(\frac{{{t_n}}}{{{t_{n + 1}}}} = \frac{{\frac{{{{\left( { - 1} \right)}^{n - 1}} \cdot {{\left( x \right)}^n}}}{{n\left( {x + 1} \right)}}}}{{\frac{{{{\left( { - 1} \right)}^n} \cdot {{\left( x \right)}^{n + 1}}}}{{n\left( {n + 1} \right)}}}}\) 

\( = \left( { - 1} \right)\frac{{n + 1}}{{n - 1}} \cdot x \cdot < 1\) 

For 0 < x < 1.

Now:

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{t_n}}}{{{t_{n + 1}}}} = \mathop {\lim }\limits_{n \to \infty } \left( { - x} \right) \cdot \frac{{1 + \frac{1}{n}}}{{1 - \frac{1}{n}}}\) 

= -x < 1.

Hence, Series S₂ is also converging in nature.