Power Series
Solved Problems
Example 3.
Find the radius of convergence and interval of convergence of the series
\[\frac{x}{1} + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + \ldots + \frac{{{x^n}}}{n} + \ldots\]
Solution.
Here \({a_n} = {\frac{1}{n}}\) and \({a_{n + 1}} = {\frac{1}{{n + 1}}}.\) Calculate the radius of convergence:
\[R = \lim\limits_{n \to \infty } \left| {\frac{{{a_n}}}{{{a_{n + 1}}}}} \right| = \lim\limits_{n \to \infty } \frac{{\frac{1}{n}}}{{\frac{1}{{n + 1}}}} = \lim\limits_{n \to \infty } \frac{{n + 1}}{n} = \lim\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right) = 1.\]
When \(x = -1,\) we have the convergent series \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}}.\) When \(x = 1,\) we obtain the divergent harmonic series \(\sum\limits_{n = 1}^\infty {\frac{1}{n}}.\) Thus, the initial series converges in the half-open interval \(\left[ { -1, 1} \right).\)
Example 4.
For what values of \(x\) does the series \[\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n + 1}}}\] converge?
Solution.
Determine the radius and interval of convergence of the series.
\[R = \lim\limits_{n \to \infty } \left| {\frac{{{a_n}}}{{{a_{n + 1}}}}} \right| = \lim\limits_{n \to \infty } \frac{{\frac{1}{{n + 1}}}}{{\frac{1}{{n + 2}}}} = \lim\limits_{n \to \infty } \frac{{n + 2}}{{n + 1}} = \lim\limits_{n \to \infty } \frac{{\frac{\cancel{n}}{\cancel{n}} + \frac{2}{n}}}{{\frac{\cancel{n}}{\cancel{n}} + \frac{1}{n}}} = \lim\limits_{n \to \infty } \frac{{1 + \frac{2}{n}}}{{1 + \frac{1}{n}}} = 1.\]
If \(x = -1,\) we get the series
\[\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n + 1}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n + 1}}} ,\]
that converges by the alternating series test.
If \(x = 1,\) we get the divergent series:
\[\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n + 1}}} = \sum\limits_{n = 0}^\infty {\frac{1}{{n + 1}}} .\]
Thus, the interval of convergence of the given series is \(\left[ { -1, 1} \right).\)
Example 5.
Find the radius of convergence and interval of convergence of the power series \[\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( {x - 2} \right)}^n}}}{{{n^2}}}}.\]
Solution.
We make the substitution: \(u = x - 2.\) The series then becomes \(\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{u^n}}}{{{n^2}}}}.\) Calculate the radius of convergence:
\[R = \lim\limits_{n \to \infty } \left| {\frac{{{a_n}}}{{{a_{n + 1}}}}} \right| = \lim\limits_{n \to \infty } \frac{{\frac{1}{{{n^2}}}}}{{\frac{1}{{{{\left( {n + 1} \right)}^2}}}}} = \lim\limits_{n \to \infty } \frac{{{{\left( {n + 1} \right)}^2}}}{{{n^2}}} = \lim\limits_{n \to \infty } {\left( {\frac{{n + 1}}{n}} \right)^2} = \lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^2} = 1.\]
Investigate convergence at the endpoints of the interval.
If \(u = -1,\) then the series
\[\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{u^n}}}{{{n^2}}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( { - 1} \right)}^n}}}{{{n^2}}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{2n}}}}{{{n^2}}}} = \sum\limits_{n = 0}^\infty {\frac{1}{{{n^2}}}} \]
converges as a \(p\)-series with \(p = 2 \gt 1.\)
If \(u = 1,\) we get the alternating series
\[\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{u^n}}}{{{n^2}}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{1^n}}}{{{n^2}}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}},\]
that converges by Leibniz's theorem.
Thus, the interval of convergence of the series \(\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{u^n}}}{{{n^2}}}}\) is \(\left[ { -1, 1} \right].\) Since \(u = x - 2,\) the initial series converges for
\[- 1 \le x - 2 \le 1\;\;\text{or}\;\;1 \le x \le 3.\]
Answer: the given series converges in the interval \(\left[ {1, 3} \right].\)
Example 6.
Find the radius of convergence and interval of convergence of the power series
\[1 + \frac{{2x}}{{\sqrt {5 \cdot 5} }} + \frac{{4{x^2}}}{{\sqrt {9 \cdot {5^2}} }} + \frac{{8{x^3}}}{{\sqrt {13 \cdot {5^3}} }} + \ldots\]
Solution.
The \(n\)th term of the series (starting from \(n = 0\)) is
\[{w_n} = {a_n}{x^n} = \frac{{{2^n}{x^n}}}{{\sqrt {\left( {4n + 1} \right) \cdot {5^n}} }}.\]
Here
\[{a_n} = \frac{{{2^n}}}{{\sqrt {\left( {4n + 1} \right) \cdot {5^n}} }} \;\;\text{and}\;\; a_{n + 1} = \frac{{{2^{n + 1}}}}{{\sqrt {\left( {4n + 5} \right) \cdot {5^{n + 1}}} }}.\]
Determine the radius of convergence:
\[R = \lim\limits_{n \to \infty } \left| {\frac{{{a_n}}}{{{a_{n + 1}}}}} \right| = \lim\limits_{n \to \infty } \left| {\frac{{\frac{{{2^n}}}{{\sqrt {\left( {4n + 1} \right) \cdot {5^n}} }}}}{{\frac{{{2^{n + 1}}}}{{\sqrt {\left( {4n + 5} \right) \cdot {5^{n + 1}}} }}}}} \right| = \lim\limits_{n \to \infty } \frac{{{2^n}}}{{{2^{n + 1}}}}\sqrt {\frac{{\left( {4n + 5} \right) \cdot {5^{n + 1}}}}{{\left( {4n + 1} \right) \cdot {5^n}}}} = \lim\limits_{n \to \infty } \frac{1}{2}\sqrt {\left( {1 + \frac{4}{{4n + 1}}} \right) \cdot 5} = \frac{{\sqrt 5 }}{2}.\]
Now we investigate convergence of the power series at the endpoints.
If \(x = -{\frac{{\sqrt 5 }}{2}},\) we get
\[\sum\limits_{n = 0}^\infty {\frac{{{2^n}{x^n}}}{{\sqrt {\left( {4n + 1} \right) \cdot {5^n}} }}} = \sum\limits_{n = 0}^\infty {\frac{{{2^n}{{\left( { - \frac{{\sqrt 5 }}{2}} \right)}^n}}}{{\sqrt {\left( {4n + 1} \right) \cdot {5^n}} }}} = \sum\limits_{n = 0}^\infty {\frac{{\cancel{2^n}{{\left( { - 1} \right)}^n}\cancel{{\left( {\sqrt 5 } \right)}^n}}}{{\cancel{2^n}\sqrt {\left( {4n + 1} \right)} \cancel{\sqrt {{5^n}}} }}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt {\left( {4n + 1} \right)} }}} .\]
This series converges by the alternating series test (or Leibniz's theorem).
If \(x = {\frac{{\sqrt 5 }}{2}},\) we have
\[\sum\limits_{n = 0}^\infty {\frac{{{2^n}{x^n}}}{{\sqrt {\left( {4n + 1} \right) \cdot {5^n}} }}} = \sum\limits_{n = 0}^\infty {\frac{{{2^n}{{\left( {\frac{{\sqrt 5 }}{2}} \right)}^n}}}{{\sqrt {\left( {4n + 1} \right) \cdot {5^n}} }}} = \sum\limits_{n = 0}^\infty {\frac{{\cancel{2^n}\cancel{{\left( {\sqrt 5 } \right)}^n}}}{{\cancel{2^n}\sqrt {\left( {4n + 1} \right)} \cancel{\sqrt {{5^n}}} }}} = \sum\limits_{n = 0}^\infty {\frac{1}{{\sqrt {\left( {4n + 1} \right)} }}} .\]
Apply the integral test:
\[
\int\limits_0^\infty {\frac{{dx}}{{\sqrt {4x + 1} }}}
= \lim\limits_{n \to \infty } \int\limits_0^n {\frac{{dx}}{{\sqrt {4x + 1} }}}
= \frac{1}{2}\lim\limits_{n \to \infty } \int\limits_0^n {\frac{{d\left( {4x + 1} \right)}}{{2\sqrt {4x + 1} }}}
= \frac{1}{2}\lim\limits_{n \to \infty } \left[ {\left. {\left( {\sqrt {4x + 1} } \right)} \right|_0^n} \right]
= \frac{1}{2}\lim\limits_{n \to \infty } \left[ {\sqrt {4n + 1} - 1} \right] = \infty .\]
We see that the series \(\sum\limits_{n = 0}^\infty {\frac{1}{{\sqrt {4n + 1} }}} \) diverges. Therefore, the interval of convergence of the initial series is \(\left[ { - {\frac{{\sqrt 5 }}{2}}, {\frac{{\sqrt 5 }}{2}}} \right).\)
