The series \(\mathop \sum \limits_{n = 1}^\infty \sqrt {\frac{{{5^n} + 1}}{{{3^n} - 1}}} \) is

Concept:

The comparison test in the limit form,

Consider two positive term series \(\mathop \sum \limits_{n = 1}^\infty {a_n}\) and \(\mathop \sum \limits_{n = 1}^\infty {b_n}\) such that \(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = L\) (finite), then both the series either converge or diverge.

Calculation:

Given series is

\(\mathop \sum \limits_{n = 1}^\infty \sqrt {\frac{{{5^n} + 1}}{{{3^n} - 1}}} \)

⇒ \({a_n} = \sqrt {\frac{{{5^n} + 1}}{{{3^n} - 1}}} = {\left( {\frac{5}{2}} \right)^{\frac{n}{2}}}\sqrt {\frac{{1 + \frac{1}{{{5^n}}}}}{{1 - \frac{1}{{{3^n}}}}}} \)

Lets take \({b_n} = {\left( {\frac{5}{2}} \right)^{\frac{n}{2}}}\) 

By comparison test in limit form,

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_n}}}{{{b_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \sqrt {\frac{{1 + \frac{1}{{{5^n}}}}}{{1 - \frac{1}{{{3^n}}}}}} = 1\;\left( {finite} \right)\)

Therefore, both the series either converge or diverge.

The series ∑b_n is a geometric progression with r = 2.5, since r > 1, series is divergent.

As b_n is divergent, a_n is also divergent.