2.13 Power Series
An infinite series of the form \[\sum ^{\infty }_{n = 0} c_n x^n = c_1x + c_2x^2 + c_3x^3 + \cdots \cdots \cdots \hspace {0.5cm} (I)\] where the \(c_i\)’s are constants, is called a power series in \(x\). Similarly, an infinite series of the form \[\sum ^{\infty }_{n = 0} c_n (x-a)^n =c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots \cdots \cdots \hspace {0.5cm} (II)\] is called a power series in \((x -a)\).
Interval of Convergence of a Power series
The totally of values of \(x\) for which a power series converges is called its interval of convergence. If there are values of \(x\) for which a power series \((I)\) or \((II)\) converges then it converges either for all values of \(x\) or for all values of \(x\) on some interval (closed, open or half open) having midpoint \(x = 0\) for \((I)\) or \(x=a\) for \((II)\).
Find the interval of convergence of the series \[x - \frac {1}{2}x^2 + \frac {1}{3}x^3 - \frac {1}{4}x^4 + \cdots \cdots \cdots + (-1)^{n-1} \hspace {0.1cm} \frac {1}{n}x^n + \cdots \cdots \]
\[\text {i.e}\hspace {0.5cm} \sum ^{\infty }_{n = 1}\frac {1}{n}x^n \]
Solution.
We use the ration test of absolutely convergent series.
\[a_n= \dfrac {x^n}{n}\hspace {0.5cm}\implies \hspace {0.5cm} a_{n + 1} = \frac {x^{n+1}}{n + 1}\]
\[\left |\dfrac {a_{n+1}}{a_n}\right | = \left |\dfrac {x^{n+ 1}}{n+ 1}\cdot \dfrac {n}{x^n}\right | = \dfrac {n}{n + 1}\hspace {0.1cm}\left |x\right |\]
\[\lim _{n\rightarrow \infty } \dfrac {n}{n + 1}\hspace {0.1cm}\left |x\right | = \left |x\right |\lim _{n\rightarrow \infty } \dfrac {n}{n + 1} = \left |x\right |\]
\[-1 < x<1\]
We must test the end points. For \(x = 1\), the series becomes \[1 - \dfrac {1}{2} + \dfrac {1}{3} - \dfrac {1}{4} + \cdots \cdots \cdots \] The series converges as an alternating series.
For \(x = -1\), the series \[-1 - \dfrac {1}{2} - \dfrac {1}{3} - \frac {1}{4}\cdots \cdots \cdots \] which can be written as \(\displaystyle {-\sum ^{\infty }_{n=1} \frac {1}{n}}\) which is a harmonic series and diverges.
\(\therefore \hspace {0.3cm}\) the interval of convergence of the given series is \(-1< x\leq 1\).
Find the interval of convergence of the series \[\dfrac {x - 2}{1} + \frac {(x - 2)^2}{2} + \dfrac {(x - 2)^3}{3} + \cdots \cdots \cdots + \frac {(x - 2)^n}{n} + \cdots \cdots \cdots \]
Solution.
Here \(\displaystyle {a_n = \frac {(x-2)^n}{n}\hspace {0.5cm} \implies \hspace {0.5cm} a_{n+1} = \frac {(x -2^{n+1}}{n+1}}\)
By the ratio test of absolutely convergent series, we have
\[\left |\dfrac {a_{n+1}}{a_n}\right | = \left |\dfrac {(x-2)^{n+1}}{n+1}\cdot \dfrac {n}{(x -2)}\right |= \dfrac {n}{n + 1}\hspace {0.1cm}\left |x - 2\right |\]
Thus \(\displaystyle {\lim \limits _{n \rightarrow \infty }\left |\dfrac {a_{n+1}}{a_n}\right |= \left |x - 2\right |\hspace {0.1cm}\lim \limits _{n \rightarrow \infty }\dfrac {n}{n + 1}= \left |x - 2\right |}\)
The series converges when \(\displaystyle {\lim \limits _{n \rightarrow \infty }\left |\dfrac {a_{n+1}}{a_n}\right | < 1}\hspace {0.3cm}\) i.e
\[\lim \limits _{n \rightarrow \infty }\left |\dfrac {a_{n+1}}{a_n}\right | = \left |x - 2\right | <1\]
i.e The series converges for \(1 < x < 3\).
Now, we need to test the end points.
For \(x =1\), the series becomes \[-1 + \dfrac {1}{2} - \dfrac {1}{3} + \dfrac {1}{4} + \cdots \cdots + \dfrac {(-1)^n}{n} + \cdots \cdots = \sum ^{\infty }_{n= 1} \dfrac {(-1)^n}{n}\] which converges by alternating series test.
For \(x = 3\), the series becomes \[1 + \dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} +\cdots \cdots \cdots = \sum ^{\infty }_{n=1} \frac {1}{n}\] which diverges.
\(\therefore \) the interval of convergence is \(1\leq x < 3\).
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