Calculus

Infinite Sequences and Series

The Ratio and Root Tests

Page 1
Problems 1-2
Page 2
Problems 3-7

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 on problem description to see 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.

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
Page 2
Problems 3-7