The Ratio and Root Tests
Solved Problems
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.
Solution.
According to the ratio test, we calculate the following limit:
\[\lim\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \lim\limits_{n \to \infty } \frac{{\frac{{{{\left( {\left( {n + 1} \right)!} \right)}^2}}}{{\left( {2\left( {n + 1} \right)} \right)!}}}}{{\frac{{{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}}}} = \lim\limits_{n \to \infty } \Big[ {\frac{{{{\left( {\left( {n + 1} \right)!} \right)}^2}}}{{{{\left( {n!} \right)}^2}}} \cdot \frac{{\left( {2n} \right)!}}{{\left( {2n + 2} \right)!}}} \Big] = \lim\limits_{n \to \infty } \Big[ {\frac{{{{\left( {n!\left( {n + 1} \right)} \right)}^2}}}{{{{\left( {n!} \right)}^2}}} \cdot \frac{\cancel{{\left( {2n} \right)!}}}{{\cancel{\left( {2n} \right)!}\left( {2n + 1} \right)\left( {2n + 2} \right)}}} \Big] = \lim\limits_{n \to \infty } \Big[ {\frac{{\cancel{{\left( {n!} \right)}^2}{{\left( {n + 1} \right)}^2}}}{{\cancel{{\left( {n!} \right)}^2}}} \cdot \frac{1}{{\left( {2n + 1} \right)\left( {2n + 2} \right)}}} \Big] = \lim\limits_{n \to \infty } \frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2n + 1} \right)\left( {2n + 2} \right)}} = \lim\limits_{n \to \infty } \frac{{{n^2} + 2n + 1}}{{4{n^2} + 2n + 4n + 2}} = \lim\limits_{n \to \infty } \frac{{{n^2} + 2n + 1}}{{4{n^2} + 6n + 2}}.\]
Divide the numerator and denominator by \({n^2}:\)
\[\lim\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \lim\limits_{n \to \infty } \frac{{\frac{{{n^2} + 2n + 1}}{{{n^2}}}}}{{\frac{{4{n^2} + 6n + 2}}{{{n^2}}}}} = \lim\limits_{n \to \infty } \frac{{1 + \frac{2}{n} + \frac{1}{{{n^2}}}}}{{4 + \frac{6}{n} + \frac{2}{{{n^2}}}}} = \frac{1}{4} \lt 1.\]
Hence, the series converges.
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.
Solution.
We apply the ratio test and calculate the corresponding limit:
\[\lim\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \lim\limits_{n \to \infty } \frac{{\frac{{{3^{n + 1}}\left( {n + 1} \right)!}}{{{{\left( {n + 1} \right)}^{n + 1}}}}}}{{\frac{{{3^n}n!}}{{{n^n}}}}} = \lim\limits_{n \to \infty } \left[ {\frac{{{3^{n + 1}}\left( {n + 1} \right)!}}{{{3^n}n!}} \cdot \frac{{{n^n}}}{{{{\left( {n + 1} \right)}^{n + 1}}}}} \right] = \lim\limits_{n \to \infty } \left[ {\frac{{3\cancel{n!}\cancel{\left( {n + 1} \right)}}}{\cancel{n!}} \cdot \frac{{{n^n}}}{{{{\left( {n + 1} \right)}^n}\cancel{\left( {n + 1} \right)}}}} \right] = 3\lim\limits_{n \to \infty } \frac{{{n^n}}}{{{{\left( {n + 1} \right)}^n}}} = 3\lim\limits_{n \to \infty } {\left( {\frac{n}{{n + 1}}} \right)^n} = \frac{3}{{\lim\limits_{n \to \infty } {{\left( {\frac{{n + 1}}{n}} \right)}^n}}} = \frac{3}{{\lim\limits_{n \to \infty } {{\left( {1 + \frac{1}{n}} \right)}^n}}} = \frac{3}{e}.\]
Since the ratio \(\frac{3}{e}\) is greater than 1, the given series is divergent.
Example 5.
Investigate whether the series \[\sum\limits_{n = 1}^\infty {\frac{{{n^n}}}{{{2^{3n - 1}}}}}\] converges or diverges.
Solution.
We use the root test.
\[\lim\limits_{n \to \infty } \sqrt[n]{{\frac{{{n^n}}}{{{2^{3n - 1}}}}}} = \lim\limits_{n \to \infty } \frac{n}{{{2^{3 - \frac{1}{n}}}}} = \lim\limits_{n \to \infty } \frac{n}{{{2^3} \cdot {2^{ - \frac{1}{n}}}}} = \lim\limits_{n \to \infty } \frac{{n \cdot {2^{\frac{1}{n}}}}}{8} = \lim\limits_{n \to \infty } \frac{n}{8} \cdot \lim\limits_{n \to \infty } \sqrt[n]{2} = \infty \cdot 1 = \infty \gt 1.\]
We see that the given series is divergent.
Example 6.
Determine whether the series \[\sum\limits_{n = 1}^\infty {{{\left( {\frac{1}{3}} + {\frac{1}{n}} \right)}^n}}\] converges or diverges.
Solution.
Using the root test, we calculate the following limit:
\[\lim\limits_{n \to \infty } \sqrt[n]{{{{\left( {\frac{1}{3} + \frac{1}{n}} \right)}^n}}} = \lim\limits_{n \to \infty } \left( {\frac{1}{3} + \frac{1}{n}} \right) = \frac{1}{3} + 0 = \frac{1}{3} \lt 1.\]
Hence, the given series converges.
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.
Solution.
The \(n\)th term of the series is \({a_n} = {\frac{1}{{{{\left( {\ln n} \right)}^n}}}}.\) Use the root test:
\[\lim\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = \lim\limits_{n \to \infty } \sqrt[n]{{\frac{1}{{{{\left( {\ln n} \right)}^n}}}}} = \lim\limits_{n \to \infty } \frac{1}{{\ln n}} = 0 \lt 1.\]
Since the limit is less than \(1,\) the series is convergent.
