# Root Test for for Convergence of an Infinite Series

## Key Questions

• I would use Root Test when the terms of the series are in the form of some expression to the nth power; otherwise, I would try other tests first.

Example

Let us look at examine the convergence of the series:

${\sum}_{n = 1}^{\infty} {\left(\frac{2 n}{5 - 3 n}\right)}^{n}$

By Root Test,

${\lim}_{n \to \infty} \sqrt[n]{| {\left(\frac{2 n}{5 - 3 n}\right)}^{n} |} = {\lim}_{n \to \infty} | \frac{2 n}{5 - 3 n} |$

by dividing the numerator and the denominator by $n$,

$= {\lim}_{n \to \infty} | \frac{2}{\frac{5}{n} - 3} | = | \frac{2}{0 - 3} | = \frac{2}{3} < 1$

Hence, the series is absolutely convergent.

I hope that this was helpful.

• Let ${a}_{n} = {\left(\frac{{n}^{2} + 1}{2 {n}^{2} + 1}\right)}^{n}$.

By Root Test,

${\lim}_{n \to \infty} \sqrt[n]{| {a}_{n} |} = {\lim}_{n \to \infty} \sqrt[n]{| {\left(\frac{{n}^{2} + 1}{2 {n}^{2} + 1}\right)}^{n} |}$

by cancelling out the nth-root and the nth-power,

$= {\lim}_{n \to \infty} \frac{{n}^{2} + 1}{2 {n}^{2} + 1}$

(Note: the absolute value is not necessary since it is already positive.)

by dividing by ${n}^{2}$,

$= {\lim}_{n \to \infty} \frac{1 + \frac{1}{n} ^ 2}{2 + \frac{1}{n} ^ 2} = \frac{1 + 0}{2 + 0} = \frac{1}{2} < 1$

Hence, the series converges.

I hope that this was helpful.

• Root Test

If ${\lim}_{n \to \infty} \sqrt[n]{| {a}_{n} |} < 1$, then ${\sum}_{n = 1}^{\infty} {a}_{n}$ converges.
If ${\lim}_{n \to \infty} \sqrt[n]{| {a}_{n} |} > 1$, then ${\sum}_{n = 1}^{\infty} {a}_{n}$ diverges.
If ${\lim}_{n \to \infty} \sqrt[n]{| {a}_{n} |} = 1$, then it is inconclusive.

I hope that this was helpful.