Question

Series \(\mathop \sum \limits_{n = 1}^\infty {\left( {\;{{\left( {\frac{{n + 1\;}}{n}} \right)}^{n + 1}} - \frac{{n + 1}}{n}} \right)^{ - n}}\) is
1. divergent
2. convergent
3. oscillates finitely
4. oscillates infinitely

Answer 1

Correct Answer - Option 2 : convergent

Concept:

Cauchy’s Root test-

A series of positive terms ∑u_nis

(i) convergent if \(\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^{\frac{1}{n}}} < 1\)

(ii) Divergent if \(\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^{\frac{1}{n}}} > 1\)

(iii) Test fail if \(\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^{\frac{1}{n}}} = 1\)

Calculation:

Given:

\(\mathop \sum \limits_{n = 1}^\infty {\left( {\;{{\left( {\frac{{n + 1\;}}{n}} \right)}^{n + 1}} - \frac{{n + 1}}{n}} \right)^{ - n}}\)

\(\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } {\left[ {{{\left( {\frac{{n + 1}}{n}} \right)}^{n + 1}} - \frac{{n + 1}}{n}} \right]^{ - 1}}\)

\(= \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{n + 1}}{n}} \right){\left[ {{{\left( {\frac{{n + 1}}{n}} \right)}^n} - 1} \right]^{ - 1}}\)

\(= \mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right){\left[ {{{\left( {1 + \frac{1}{n}} \right)}^n} - 1} \right]^{ - 1}}\)

= (e – 1)^-1

\(\frac{1}{{e - 1}}\)

\(e≃2.7\)

\(= \frac{1}{{1.7}} < 1\)

Hence, it is a convergent series.