Informative line

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

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 = $$(-1,1)$$  #### 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 = $$(-1,1)$$

#### 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

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 = $$(-1,1)$$

#### 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 = $$(-1,1)$$

#### 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