Question

Use the Root Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}(2 \sqrt[n]{n}+1)^{n} $$

Short answer

The series \(\sum_{n=1}^{\infty}(2 \sqrt[n]{n}+1)^{n}\) is divergent.

Step-by-step solution

Apply the Root Test

The expression of which we need to find the nth root is \( |(2 \sqrt[n]{n}+1)^{n}| \). This simplifies to \( |2 \sqrt[n]{n} + 1| \) because the nth root and the nth power cancel each other out.

Calculate the limit

Find \(\lim_{n \to \infty} (|2 \sqrt[n]{n} + 1|)\). As n approaches infinity, \( \sqrt[n]{n} \) approaches 1. So, \(\lim_{n \to \infty} (|2 \sqrt[n]{n} + 1|) = |2*1 + 1| = 3\)

Determine convergence or divergence

By the Root Test, the series is convergent if the limit found is less than 1, divergent if it's greater than 1, and the test is inconclusive if it equals 1. Here, since the limit is 3 which is greater than 1, we can conclude that the series \(\sum_{n=1}^{\infty}(2 \sqrt[n]{n}+1)^{n}\) is divergent.