Question

Use the Root Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(\frac{4 n+3}{2 n-1}\right)^{n} $$

Short answer

The given series is divergent.

Step-by-step solution

Apply the Root Test

To apply the Root Test to our series, compute the limit of the nth root of the absolute value of the nth term of the series as n approaches infinity. For our series, the nth term is \((\frac{4n+3}{2n-1})^n\). Therefore, we compute the limit: \[ \lim_{{n\to\infty}}\sqrt[n]{{\left|\left(\frac{4n+3}{2n-1}\right)^n\right|}} \]

Simplify the limit

The nth root of a term raised to the power of n simplifies the expression. So the limit simplifies to:\[ \lim_{{n\to\infty}}\left|\frac{4n+3}{2n-1}\right| \]

Evaluate the limit

When taking limit of a rational function where the highest power of n in the numerator and denominator is the same, divide every term by \(n^{highest power}\). Here it is 1. So:\[\lim_{{n\to\infty}}\left|\frac{4+\frac{3}{n}}{2-\frac{1}{n}}\right| \]This simplifies as n goes to infinity, to:\[\left|\frac{4}{2}\right| = 2\]

Determine convergence or divergence

As 2 is greater than 1, according to the Root Test, the series is divergent.