# 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 $$\lim\limits_{n \to \infty } {\large\frac{{{a_{n + 1}}}}{{{a_n}}}\normalsize} \lt 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is convergent;
• If $$\lim\limits_{n \to \infty } {\large\frac{{{a_{n + 1}}}}{{{a_n}}}\normalsize} \gt 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is divergent;
• If $$\lim\limits_{n \to \infty } {\large\frac{{{a_{n + 1}}}}{{{a_n}}}\normalsize} = 1,$$ 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 $$\lim\limits_{n \to \infty } \sqrt[\large n\normalsize]{{{a_n}}} \lt 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is convergent;
• If $$\lim\limits_{n \to \infty } \sqrt[\large n\normalsize]{{{a_n}}} \gt 1,$$ then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is divergent;
• If $$\lim\limits_{n \to \infty } \sqrt[\large n\normalsize]{{{a_n}}} = 1,$$ 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 a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 3

Determine whether the series
${\frac{{{{\left( {1!} \right)}^2}}}{{2!}} + \frac{{{{\left( {2!} \right)}^2}}}{{4!}} + \ldots }+{ \frac{{{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}} + \ldots}$
converges or diverges.

### Example 4

Determine whether the series
${\frac{{3 \cdot 1!}}{1} + \frac{{{3^2} \cdot 2!}}{{{2^2}}} }+{ \frac{{{3^3} \cdot 3!}}{{{3^3}}} + \ldots }+{ \frac{{{3^n} \cdot n!}}{{{n^n}}} + \ldots}$
converges or diverges.

### Example 5

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

### Example 6

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

### Example 7

Investigate whether the series
${\frac{1}{{{{\left( {\ln 2} \right)}^2}}} + \frac{1}{{{{\left( {\ln 3} \right)}^3}}} }+{ \frac{1}{{{{\left( {\ln 4} \right)}^4}}} + \ldots }$
converges or diverges.

### Example 1.

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

Solution.

We use the ratio test.

${\lim\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} } = {\lim\limits_{n \to \infty } \frac{{\frac{{{3^{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}}}{{\frac{{{3^n}}}{{{n^2}}}}} } = {\lim\limits_{n \to \infty } \left[ {\frac{{{3^{n + 1}}}}{{{3^n}}} \cdot }\right.}\kern0pt{\left.{ \frac{{{n^2}}}{{{{\left( {n + 1} \right)}^2}}}} \right] } = {\lim\limits_{n \to \infty } \left[ {3{{\left( {\frac{n}{{n + 1}}} \right)}^2}} \right] } = {3\lim\limits_{n \to \infty } {\left( {\frac{{n + 1 – 1}}{{n + 1}}} \right)^2} } = {3\lim\limits_{n \to \infty } {\left( {1 – \frac{1}{{n + 1}}} \right)^2} }={ 3.}$

As it can be seen, the given series diverges.

### Example 2.

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

Solution.

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

${\lim\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} } = {\lim\limits_{n \to \infty } \frac{{\frac{{{{\left( {n + 1} \right)}^3}}}{{{{\left( {\ln 3} \right)}^{n + 1}}}}}}{{\frac{{{n^3}}}{{{{\left( {\ln 3} \right)}^n}}}}} } = {\lim\limits_{n \to \infty } \left[ {\frac{{{{\left( {\ln 3} \right)}^n}}}{{{{\left( {\ln 3} \right)}^{n + 1}}}} \cdot }\right.}\kern0pt{\left.{ \frac{{{{\left( {n + 1} \right)}^3}}}{{{n^3}}}} \right] } = {\lim\limits_{n \to \infty } \left[ {\frac{1}{{\ln 3}} \cdot }\right.}\kern0pt{\left.{ {{\left( {\frac{{n + 1}}{n}} \right)}^3}} \right] } = {\frac{1}{{\ln 3}}\lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^3} } = {\frac{1}{{\ln 3}} \cdot 1 }={ \frac{1}{{\ln 3}}.}$

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

Page 1
Problems 1-2
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Problems 3-7