Root Test for Series: Statement, Solved Examples

Cauchy’s Root test for series is used to test the convergence or divergence of an infinite series. It states that if lim_n→∞ a_n^1/n is less than 1 then the series ∑a_n converges and if the limit is greater than 1 then the series diverges. In this article, we will study Cauchy root test of a series with examples.

Root Test Rule

Statement: Let ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ be a series of positive real numbers. Suppose that

R = $\underset{n\to \mathrm{\infty }}{lim}(a_n{)}^{1/n}$.

Then by root test we have the following.

• If R<1, then the series ∑a_n converges.
• If R>1, then ∑a_n diverges.
• If R=1, the root test is indecisive.

Root Test Solved Examples

$\boxed{{\displaystyle \mathbf{\text{Q}}1:}}$ Test the convergence of $1+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{3^3}}+\cdots$.

Answer:

The given series = ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^n}}}$

Here, $a_n={\displaystyle \frac{1}{n^n}}$.

Therefore,

R = lim_n→∞ $(a_n{)}^{1/n}$

= lim_n→∞ ${\left({\displaystyle \frac{1}{n^n}}\right)}^n$

= lim_n→∞ ${\displaystyle \frac{1}{n}}$

= 0.

Since the limit R<1, the given series converges by root test.

$\boxed{{\displaystyle \mathbf{\text{Q}}2:}}$ Test the convergence of the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ where a_n = ${\displaystyle \frac{(1+n{)}^{n^2}}{n^{n^2}}}$.

Answer:

R = lim_n→∞ $(a_n{)}^{1/n}$

= lim_n→∞ ${\left({\displaystyle \frac{(1+n{)}^{n^2}}{n^{n^2}}}\right)}^n$

= lim_n→∞ ${\left({\displaystyle \frac{n+1}{n}}\right)}^n$

= lim_n→∞ ${(1+{\displaystyle \frac{1}{n}})}^n$

= e >1 as the value of e lies between 2 and 3.

Since the limit R>1, the given series diverges by root test.

Related Articles:

• Convergence of a Series: Definition, Formula, Examples
• Comparison Test for Series: Statement, Examples [Limit Form]
• Ratio Test for Series: Statement, Solved Examples

Homework:

Test the convergence of the following series:

1. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{n^{n^2}}{(1+n{)}^{n^2}}}}$
2. ${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3^2}}+{\displaystyle \frac{1}{4^3}}+\cdots$

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