Informative line

Radius Of Convergence

Find the radius of convergence and interval of convergence for the power series.

Radius of Convergence

 For a given power series   \(\sum\limits_{n=0}^\infty C_n (x-a)^n\) , there are only three possibilities .

(1) The series converges only when  \(x = a\) .

(2) The series converges for all  \(x \) .

(3) There is a positive number R such that the series converges if \(|x-a|<R\) and diverges if \(|x-a|>R\) . 

  • The number R is called the radius of convergence of power series.
  • In (1) the value of \(R=0\)
  • In (2) the value of  \(R= \infty\).
  • The interval of convergence of power series is the interval that consists of  values of   \(x \)  for which series converges .
  • In (1) the interval is a single point a.
  • In (2) the interval is \((-\infty,\infty)\)
  • In (3) there are four possibilities about interval of convergence. 

        \((a-R,\; a+R),\,\, [a-R,\; a+R), \,\,(a-R, \;a+R],\,\,[a-R, \;a+R]\) 

e.g. (1 ) \(\sum\limits_{n=0}^\infty x^n\) has interval of convergence as \((-1,1)\) and radius of convergence =1

\(\sum\limits_{n=0}^\infty x^n = 1+x+x^2+x^3+.....x^n\)

Now  \(a_n = x^n = \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{x^{n+1}}{x^n}\right| = |n|\)

For convergence      \(\left|\dfrac{a_{n+1}}{a_n}\right|<1\)

\(\Rightarrow |x| <1 \Rightarrow x\in(-1,1)\)

\(\therefore \) radius of convergence =1

and interval of convergence = \((-1,1)\)

Illustration Questions

Find the radius of convergence for the following  series \(\sum\limits_{n=1}^\infty \left(\dfrac{3^n(x+4)^n}{\sqrt n}\right)\)

A \(R= \dfrac{1}{4}\)

B \(R= \dfrac{1}{3}\)

C \(R=3\)

D \(R=4\)

×

Apply Ratio test for convergence

 

In this case \(a_n = \dfrac{3^n(x+4)^n}{\sqrt n}\)

\(\therefore \left|\dfrac{a_{n+1}}{a_n}\right| =\left|\dfrac{3^{n+1}(x+4)^{n+1}}{\dfrac{\sqrt{n+1}}{\dfrac{3^n(x+4)^n}{\sqrt n}}}\right| = 3|x+4| × \dfrac{\sqrt n}{\sqrt{n+1}}\)

\(= 3|x+4| \sqrt{\dfrac{1}{1+\dfrac{1}{n}}}\)

\(\therefore \lim\limits_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim\limits_{n\to\infty}3|x+4| × \sqrt{\dfrac{1}{1+\dfrac{1}{n}}} = 3|x+4|\)

for convergence \(3|x+4| <1 \Rightarrow |x+4| <\dfrac{1}{3}\)  

\(\Rightarrow \dfrac{-1}{3} <x+4 <\dfrac{1}{3}\)

\(\Rightarrow \dfrac{-13}{3} <x <\dfrac{-11}{3}\)

\(\therefore \) Interval  of convergence is \(\left( \dfrac{-13}{3} ,\dfrac{-11}{3}\right)\) and

radius of convergence = \(\dfrac{1}{3}\)

Find the radius of convergence for the following  series \(\sum\limits_{n=1}^\infty \left(\dfrac{3^n(x+4)^n}{\sqrt n}\right)\)

A

\(R= \dfrac{1}{4}\)

.

B

\(R= \dfrac{1}{3}\)

C

\(R=3\)

D

\(R=4\)

Option B is Correct

Interval of Convergence

 For a given power series   \(\sum\limits_{n=0}^\infty C_n (x-a)^n\) , there are only three possibilities .

(1) The series converges only when  \(x = a\) .

(2) The series converges for all  \(x \) .

(3) There is a positive number R such that the series converges if \(|x-a|<R\) and diverges if \(|x-a|>R\) . 

  • The number R is called the radius of convergence of power series.
  • In (1) the value of \(R=0\)
  • In (2) the value of  \(R= \infty\).
  • The interval of convergence of power series is the interval that consists of  values of   \(x \)  for which series converges .
  • In (1) the interval is a single point a.
  • In (2) the interval is \((-\infty,\infty)\)
  • In (3) there are four possibilities about interval of convergence. 

        \((a-R,\; a+R),\,\, [a-R,\; a+R), \,\,(a-R, \;a+R],\,\,[a-R, \;a+R]\) 

e.g. (1 ) \(\sum\limits_{n=0}^\infty x^n\) has interval of convergence as \((-1,1)\) and radius of convergence =1

\(\sum\limits_{n=0}^\infty x^n = 1+x+x^2+x^3+.....x^n\)

Now  \(a_n = x^n = \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{x^{n+1}}{x^n}\right| = |n|\)

For convergence      \(\left|\dfrac{a_{n+1}}{a_n}\right|<1\)

\(\Rightarrow |x| <1 \Rightarrow x\in(-1,1)\)

\(\therefore \) radius of convergence =1

and interval of convergence = \((-1,1)\)

 

 

Illustration Questions

Find the interval of convergence of the series  \(\sum\limits_{n=1}^\infty\dfrac{n!(2x-1)^n}{(n+1)!}\)

A \((0,4)\)

B \((1,2)\)

C \((-1,3)\)

D \((0,1)\)

×

Use Ratio test of convergence

In the case , \(a_n = \dfrac{n!(2x-1)^n}{(n+1)!}\)

\(\therefore \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{\dfrac{(n+1)!(2x-1)^{n+1}}{(n+2)!}}{\dfrac{n!(2x-1)^n}{(n+1)!}}\right|\)

\(= \dfrac{(n+1)!(n+1)!}{(n+2)!\,n!}× |2x-1| = \dfrac{n+1}{n+2} |2x-1|\)

\(\therefore \lim\limits_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \lim\limits_{n\to\infty} \dfrac{n+1}{n+2} |2x-1|\)

\(= \lim\limits_{n\to\infty} \left(\dfrac{1+\dfrac{1}{n}}{1+\dfrac{2}{n}}\right) |2x-1| = |2x-1|\)

For convergence \(|2x-1|<1 \Rightarrow -1<2x-1<1\)

\(\Rightarrow 0<2x<2\)

\(\Rightarrow 0<x<1\)

\(\therefore \) interval of convergence is \((0,1)\)

Find the interval of convergence of the series  \(\sum\limits_{n=1}^\infty\dfrac{n!(2x-1)^n}{(n+1)!}\)

A

\((0,4)\)

.

B

\((1,2)\)

C

\((-1,3)\)

D

\((0,1)\)

Option D is Correct

Radius of Convergence(Complex problems)

For a given power series   \(\sum\limits_{n=0}^\infty C_n (x-a)^n\) , there are only three possibilities .

(1) The series converges only when  \(x = a\) .

(2) The series converges for all  \(x \) .

(3) There is a positive number R such that the series converges if \(|x-a|<R\) and diverges if \(|x-a|>R\) . 

  • The number R is called the radius of convergence of power series.
  • In (1) the value of \(R=0\)
  • In (2) the value of  \(R= \infty\).
  • The interval of convergence of power series is the interval that consists of  values of   \(x \)  for which series converges .
  • In (1) the interval is a single point a.
  • In (2) the interval is \((-\infty,\infty)\)
  • In (3) there are four possibilities about interval of convergence. 

        \((a-R,\; a+R),\,\, [a-R,\; a+R), \,\,(a-R, \;a+R],\,\,[a-R, \;a+R]\) 

e.g. (1 ) \(\sum\limits_{n=0}^\infty x^n\) has interval of convergence as \((-1,1)\) and radius of convergence =1

\(\sum\limits_{n=0}^\infty x^n = 1+x+x^2+x^3+.....x^n\)

Now  \(a_n = x^n = \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{x^{n+1}}{x^n}\right| = |n|\)

For convergence      \(\left|\dfrac{a_{n+1}}{a_n}\right|<1\)

\(\Rightarrow |x| <1 \Rightarrow x\in(-1,1)\)

\(\therefore \) radius of convergence =1

and interval of convergence = \((-1,1)\)

Illustration Questions

Find the radius of convergence of the series  \(\sum\limits_{n=0}^{\infty} \dfrac{(n!)^3}{(3n!)} x^n\)

A \(8\)

B \(64\)

C \(27\)

D \(16\)

×

By the ratio test, 

If \(\lim\limits_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right| = L<1\)  then the series converges

In the case, \(a_n = \dfrac{(n!)^3}{(3n!)} x^n\) 

\(\therefore \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{\dfrac{({(n+1)!})^3}{(3(n+1))!}x^{n+1}}{\dfrac{(n!)^3}{3n!}x^n}\right|\)

\(= \left|\dfrac{(n+1)^3x}{(3n+1)(3n+2)(3n+3)}\right| = \dfrac{(n+1)(n+1)(n+1)}{(3n+1)(3n+2)(3n+3)|x|}\)

As  \(n\to \infty \left|\dfrac{a_{n+1}}{a_n}\right| \to \dfrac{|x|}{27}\)

\(\therefore \dfrac{|x|}{27} <1 \) for convergence 

\(\Rightarrow |x| <27\)

\(\Rightarrow n \,\epsilon (-27,27)\)

\(\therefore \) radius of convergence = 27

interval of convergence is  \((-27,27)\)

Find the radius of convergence of the series  \(\sum\limits_{n=0}^{\infty} \dfrac{(n!)^3}{(3n!)} x^n\)

A

\(8\)

.

B

\(64\)

C

\(27\)

D

\(16\)

Option C is Correct

Interval of Convergence(Complex problems)

 For a given power series   \(\sum\limits_{n=0}^\infty C_n (x-a)^n\) , there are only three possibilities .

(1) The series converges only when  \(x = a\) .

(2) The series converges for all  \(x \) .

(3) There is a positive number R such that the series converges if \(|x-a|<R\) and diverges if \(|x-a|>R\) . 

  • The number R is called the radius of convergence of power series.
  • In (1) the value of \(R=0\)
  • In (2) the value of  \(R= \infty\).
  • The interval of convergence of power series is the interval that consists of  values of   \(x \)  for which series converges .
  • In (1) the interval is a single point a.
  • In (2) the interval is \((-\infty,\infty)\)
  • In (3) there are four possibilities about interval of convergence. 

        \((a-R,\; a+R),\,\, [a-R,\; a+R), \,\,(a-R, \;a+R],\,\,[a-R, \;a+R]\) 

e.g. (1 ) \(\sum\limits_{n=0}^\infty x^n\) has interval of convergence as \((-1,1)\) and radius of convergence =1

\(\sum\limits_{n=0}^\infty x^n = 1+x+x^2+x^3+.....x^n\)

Now  \(a_n = x^n = \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{x^{n+1}}{x^n}\right| = |n|\)

For convergence      \(\left|\dfrac{a_{n+1}}{a_n}\right|<1\)

\(\Rightarrow |x| <1 \Rightarrow x\in(-1,1)\)

\(\therefore \) radius of convergence =1

and interval of convergence = \((-1,1)\)

Illustration Questions

Find the interval of convergence of the series \(\sum\limits_{n=1}^\infty \dfrac{(2x-1)^n}{5^n \sqrt n}\)

A \((4,6)\)

B \((-4,1)\)

C \((-3,2)\)

D \((-2,3)\)

×

By the ratio test, 

If \(\lim\limits_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right| = L<1\)  then the series converges. 

In the case  \(a_n = \dfrac{(2x-1)^n}{5^n \sqrt n}\)

\(\therefore \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{\dfrac{(2n-1)^{n+1}}{5^{n+1}\sqrt{n+1}}}{\dfrac{(2n-1)^n}{5^n \sqrt n}}\right| = \dfrac{(2n-1)}{5} \dfrac{\sqrt {n+1}}{\sqrt n}\)

As \(n \to \infty \left|\dfrac{a_{n+1}}{a_n}\right| \to \left|\dfrac{2n-1}{5}\right|\)

\(\therefore \left|\dfrac{2n-1}{5}\right| <1\) for convergence

\(\Rightarrow -1 < \dfrac{2n-1}{5} <1\)

\(\Rightarrow -5 <2n -1 <5\)

\(\Rightarrow -4 <2n <6\)

\(\Rightarrow -2 <n <3\)

\(\therefore \) interval of convergence is \((-2,3)\)

Find the interval of convergence of the series \(\sum\limits_{n=1}^\infty \dfrac{(2x-1)^n}{5^n \sqrt n}\)

A

\((4,6)\)

.

B

\((-4,1)\)

C

\((-3,2)\)

D

\((-2,3)\)

Option D is Correct

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