Question

In Exercises \(63-74,\) use the Root Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(\frac{n}{2 n+1}\right)^{n} $$

Short answer

The series \(\sum_{n=1}^{\infty}\left(\frac{n}{2 n+1}\right)^{n}\) converges by the Root Test.

Step-by-step solution

Identify the sequence within the series and apply the Root Test

The sequence within the series is \(a_{n}=\left(\frac{n}{2 n+1}\right)^{n}\). The Root Test involves computing the limit \(L\) of the nth root of the sequence as \(n\) approaches infinity. That is, \(L = \lim _{n \rightarrow \infty}\sqrt[n]{|a_{n}|}\). We substitute \(a_{n}\) into the formula to get: \(L = \lim _{n \rightarrow \infty}\sqrt[n]{\left|\left(\frac{n}{2 n+1}\right)^{n}\right|}\)

Simplify the limit

The absolute value is unnecessary because the sequence is always positive. And \(n\) root of \(n\)th power will cancel out. So simplify to: \(L = \lim _{n \rightarrow \infty} \frac{n}{2 n+1}\)

Compute the limit

When \(n\) approaches infinity, the highest power of \(n\) in the denominator and the numerator will dominate the other terms. Therefore, the limit gives: \(L = \lim _{n \rightarrow \infty} \frac{n}{2 n+1} = \frac{1}{2}\)

Determine convergence or divergence

Because \(L = \frac{1}{2} < 1\), according to the Root Test, the series converges.