Question

Find Whether \(\sum\limits_{n = 1}^\infty {\sqrt[n](2)} \) Is Convergent (Or) Divergent. If It Is Convergent Find The Summation.

Short answer

Given \(\sum\limits_{n = 1}^\infty {\sqrt[n]{2}} = {\sum\limits_{n = 1}^\infty {(2)} ^{\frac{1}{n}}}\) \(\)

Hence\(\sum\limits_{n = 1}^\infty {\sqrt[n](2)} \) Diverges byDivergence Series Test

Step-by-step solution

Evaluate The Expression

\(\sum\limits_{n = 1}^\infty {\sqrt[n](2)} = {(2)^1} + {(2)^{\frac{1}{2}}} + {(2)^{\frac{1}{3}}} + {(2)^{\frac{1}{4}}} + - - - - - - \)

It is not a Geometric Series and not possible to Evaluate without Test for

Divergence.

Test For Divergence

Consider \(\sum\limits_{n = 1}^\infty {\sqrt[n](2)} = \sum\limits_{n = 1}^\infty {{{(2)}^{\frac{1}{n}}}} \)

\( \Rightarrow \mathop {\lim }\limits_{n \to \infty } {(2)^{\frac{1}{n}}} = {(2)^{\frac{1}{\infty }}}\)

\( = {(2)^0}\) \((\frac{1}{\infty } = 0)\)

= 1

The Limit of The \(\sqrt[n](2) \ne 0\)

Hence\(\sum\limits_{n = 1}^\infty {\sqrt[n](2)} \) Diverges byDivergence Series Test