Series and convergence

Hi

I'm utterly stumped by this:

Find the exact range of x for which the sum to infinity (from n = 1) of (x^n)/( (4^n)n) converges.

Any pointers would be very much appreciated.

Thanks in advance :smile:

Reply 1

ztibor

10

Original post by *Georgine*

Hi

I'm utterly stumped by this:

Find the exact range of x for which the sum to infinity (from n = 1) of (x^n)/( (4^n)n) converges.

Any pointers would be very much appreciated.

Thanks in advance :smile:

If you have s serioes of

\displaystyle \sum^{\infty}_{n=1}a_n(x-A)^n

then the radius of convergence (R) around 'A' is

\frac{1}{R}=lim_{n \to \infty}\sqrt[n]{a_n}

In your example A=0, At the limit you can use any criteria of convergence,
for example the ratio of

\frac{a_{n+1}}{a_n}

Note that

lim_{n \to \infty}\sqrt[n]{n}=1

(edited 13 years ago)

Reply 2

*Georgine*

OP

Original post by ztibor

If you have s serioes of

\displaystyle \sum^{\infty}_{n=1}a_n(x-A)^n

then the radius of convergence (R) around 'A' is

\frac{1}{R}=lim_{n \to \infty}\sqrt[n]{a_n}

In your example A=0, At the limit you can use any criteria of convergence,
for example the ratio of

\frac{a_{n+1}}{a_n}

Note that

lim_{n \to \infty}\sqrt[n]{n}=1

I really do not understand this. How does the ratio give me the exact values of x? And what would be the a (little n) in my case?

Reply 3

Daniel Freedman

7

You are considering

\displaystyle \sum_{n=1}^{\infty} a_n

, with

a_n = \frac{x^n}{4^n n}

.

By the ratio test:

- if

\displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1

then the series converges.

- if

\displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1

then the series diverges.

- if

\displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1

, then you can't tell anything and need to examine this case separately.

Use the first bullet point to find what range x has to be in to make the ratio less than 1 - you know that the series converges for all x in this range.

Then examine the case where the ratio = 1 directly i.e. find the value of x that makes the ratio 1, and see if you think the series converges/diverges for this particular value of x.

(edited 13 years ago)

Reply 4

username22878

2

Original post by Daniel Freedman

Then examine the case where the ratio = 1 directly i.e. find the value of x that makes the ratio 1, and see if you think the series converges/diverges for this particular value of x.

Given that there were no restrictions on where x lies, I would say 'values of x that make the ratio 1'.

Reply 5

Daniel Freedman

7

Original post by ljfrugn

Given that there were no restrictions on where x lies, I would say 'values of x that make the ratio 1'.

Indeed. Thanks

Reply 6

ztibor

10

Original post by *Georgine*

I really do not understand this. How does the ratio give me the exact values of x? And what would be the a (little n) in my case?

Not the ratio gives but the limit of ratio does a value which maybe finite or infinite or even 0. This value is the reciprotial of R radius,
For example if the limit is L then

\displaystyle R=\frac{1}{L}

and the series will convergent if

x \in [A-R, A+R]

For your example:

\displaystyle \sum^{\infty}_{n=1} \frac{x^n}{4^n \cdot n}

so

\displaystyle a_n=\frac{1}{4^n \cdot n}

With the nth root test

\displaystyle \frac{1}{R}=lim_{n \to \infty}\sqrt[n]{\frac{1}{4^n \cdot n}}=lim_{n \to \infty}\frac{1}{4 \cdot \sqrt[n]{n}}=\frac{1}{4}

with the ratio

\displaystyle \frac{1}{R}=lim_{n \to \infty}\frac{1}{4^{n+1} \cdot (n+1)} \cdot \frac{4^n \cdot n}{1}=lim_{n \to \infty}\frac{4^n \cdot n}{4\cdot 4^n \cdot (n+1)}=

\displaystyle =lim_{n \to \infty}\frac{1}{4 \cdot (1+\frac{1}{n})}=\frac{1}{4}

Calculating the limit with R you will get the range of convergence and consider A=0 (convergence around zero)

(edited 13 years ago)

Series and convergence

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