# Calculus

## Infinite Sequences and Series # The Ratio and Root Tests

## The Ratio Test

Let $$\sum\limits_{n = 1}^\infty {{a_n}}$$ be a series with positive terms. Then the following rules are valid:

• If $$\lim\limits_{n \to \infty } {\frac{{{a_{n + 1}}}}{{{a_n}}}} \lt 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is convergent;
• If $$\lim\limits_{n \to \infty } {\frac{{{a_{n + 1}}}}{{{a_n}}}} \gt 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is divergent;
• If $$\lim\limits_{n \to \infty } {\frac{{{a_{n + 1}}}}{{{a_n}}}} = 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ may converge or diverge and the ratio test is inconclusive; some other tests must be used.

## The Root Test

Consider again the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ with positive terms. According to the Root Test:

• If $$\lim\limits_{n \to \infty } \sqrt[n]{{{a_n}}} \lt 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is convergent;
• If $$\lim\limits_{n \to \infty } \sqrt[n]{{{a_n}}} \gt 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is divergent;
• If $$\lim\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}},$$ may converge or diverge, but no conclusion can be drawn from this test.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Determine whether the series $\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{n^2}}}}$ converges or diverges.

### Example 2

Investigate whether the series $\sum\limits_{n = 1}^\infty {\frac{{{n^3}}}{{{{\left( {\ln 3} \right)}^n}}}}$ converges or diverges.

### Example 1.

Determine whether the series $\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{n^2}}}}$ converges or diverges.

Solution.

We use the ratio test.

$\lim\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \lim\limits_{n \to \infty } \frac{{\frac{{{3^{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}}}{{\frac{{{3^n}}}{{{n^2}}}}} = \lim\limits_{n \to \infty } \left[ {\frac{{{3^{n + 1}}}}{{{3^n}}} \cdot \frac{{{n^2}}}{{{{\left( {n + 1} \right)}^2}}}} \right] = \lim\limits_{n \to \infty } \left[ {3{{\left( {\frac{n}{{n + 1}}} \right)}^2}} \right] = 3\lim\limits_{n \to \infty } {\left( {\frac{{n + 1 - 1}}{{n + 1}}} \right)^2} = 3\lim\limits_{n \to \infty } {\left( {1 - \frac{1}{{n + 1}}} \right)^2} = 3.$

As it can be seen, the given series diverges.

### Example 2.

Investigate whether the series $\sum\limits_{n = 1}^\infty {\frac{{{n^3}}}{{{{\left( {\ln 3} \right)}^n}}}}$ converges or diverges.

Solution.

We apply the ratio test to investigate convergence of this series:

$\lim\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \lim\limits_{n \to \infty } \frac{{\frac{{{{\left( {n + 1} \right)}^3}}}{{{{\left( {\ln 3} \right)}^{n + 1}}}}}}{{\frac{{{n^3}}}{{{{\left( {\ln 3} \right)}^n}}}}} = \lim\limits_{n \to \infty } \left[ {\frac{{{{\left( {\ln 3} \right)}^n}}}{{{{\left( {\ln 3} \right)}^{n + 1}}}} \cdot \frac{{{{\left( {n + 1} \right)}^3}}}{{{n^3}}}} \right] = \lim\limits_{n \to \infty } \left[ {\frac{1}{{\ln 3}} \cdot {{\left( {\frac{{n + 1}}{n}} \right)}^3}} \right] = \frac{1}{{\ln 3}}\lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^3} = \frac{1}{{\ln 3}} \cdot 1 = \frac{1}{{\ln 3}}.$

As $$\ln 3 \gt \ln e = 1$$ and $${\frac{1}{{\ln 3}}} \lt 1,$$ the given series converges.

See more problems on Page 2.