Question

How do you find the radius of convergence of the binomial power series?

Answer 1

Hint: In this question, we have to find the radius of convergence of the binomial power series. The binomial series is the expansion of $ {\left( {1 + x} \right)^n} $ . We have to find the radius of convergence of this series, so we will first apply the ratio test and then get the interval of x for which the given series converges, this interval is known as the interval of convergence. The half of the length of the interval of convergence is the radius of the convergence.

Complete step by step solution:
A binomial power series is given as –
 $ {\left( {1 + x} \right)^a} = \sum\limits_{n = 0}^\infty {^\alpha {C_n}\left( {{x^n}} \right)} $
So the nth term of this sequence is $ {\alpha _n} = \left( {^\alpha {C_n}} \right){x^n} $ where $ ^\alpha {C_n} = \dfrac{{\alpha !}}{{\left( {\alpha - n} \right)!n!}} $ .
Now we will do the ratio test of this series, we have $ L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| $
Putting the value of $ {a_n} $ in the above equation, we have –
 $ L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{\left( {^\alpha {C_{n + 1}}} \right){x^{n + 1}}}}{{\left( {^\alpha {C_n}} \right){x^n}}}} \right| $
 $ \Rightarrow L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{\dfrac{{\alpha !}}{{\left( {\alpha - \left( {n + 1} \right)} \right)!\left( {n + 1} \right)!}}{x^{n + 1}}}}{{\dfrac{{\alpha !}}{{\left( {\alpha - n} \right)!n!}}{x^n}}}} \right| $
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{\left( {\alpha - n} \right)!n!}}{{\left( {\alpha - \left( {n + 1} \right)} \right)!\left( {n + 1} \right)!}}x} \right|\]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{\left( {\alpha - n} \right)}}{{\left( {n + 1} \right)}}x} \right|\]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{\left( {\dfrac{\alpha }{n} - 1} \right)}}{{\left( {1 + \dfrac{1}{n}} \right)}}x} \right|\]
Dividing both the numerator and the denominator by “n”, we get –
As n approaches infinity, $ \left( {\dfrac{1}{n}} \right) $ approaches zero, so $ \left( {\dfrac{\alpha }{n}} \right) $ also approaches zero, so we get,
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{\left( {0 - 1} \right)}}{{\left( {1 + 0} \right)}}x} \right|\]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{ - 1}}{1}x} \right|\]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to \infty } \left| { - x} \right|\]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to \infty } \left| x \right|\]
As we are given that the series converges so \[\left| x \right| < 1\].
So, the interval of convergence is $ \left( { - 1,1} \right) $ .
The radius of convergence is $ \dfrac{{1 - \left( { - 1} \right)}}{2} = \dfrac{2}{2} = 1 $ .
Hence, the radius of convergence of the binomial power series is $ 1 $ .


Note: A test called the ratio test, also known as D’Alembert’s ratio test or the Cauchy ratio test is done to test the convergence of a series $ \sum\limits_{n = 1}^\infty {{a_n}} $ . The test is given as $ L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| $ , if the value of L comes out to be smaller than 1 then the series converges, if it is equal to 1 then the test is inconclusive as the limit fails to exist and if the value comes out to be greater than 1 then the series is divergent.