The Ratio and Root Tests

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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.