Does the Double Factorial Affect the Convergence of the Powerseries?

[Mathman23]

Hi

I'm suppose to show that the powerseries

\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1}

converge for all z \in mathbb{C}

By using the Ratio test:

I get

\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1} = \frac{2(2n+1)}{2n+3} |z^2| = \frac{2}{3}|z^2|, n \rightarrow \infty

I guess that can be re-written as |z^2| < 3/2 ?

But does that mean the ratio test fails?

Sincerley

Fred

 

[Curious3141]

Hi

I'm suppose to show that the powerseries

\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1}

converge for all z \in mathbb{C}

By using the Ratio test:

I get

\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1} = \frac{2(2n+1)}{2n+3} |z^2| = \frac{2}{3}|z^2|, n \rightarrow \infty

I guess that can be re-written as |z^2| < 3/2 ?

But does that mean the ratio test fails?

Sincerley

Fred

Firstly, are you sure that series converges for *all* z in C with no restrictions on the magnitude?

Secondly, when doing the ratio test, you should write the sequence as being equal to the ratio of two successive terms (I mean the notation is incorrect - you're effectively equating the series sum to the ratio when you put an equal sign there).

 

[Jameson]

I'm going to do the Ratio Test as I type to check your work.

Calling b_n=\frac{a_{n+1}}{a_n} I get that b_n=\frac{(2^{n+1})(z^{2n+3})}{2n+3} \times \frac{2n+1}{(2^n)(z^{2n+1})}.

Now for simplification.

b_n=\frac{2*2^n*z^{2n}*z^3}{2n+3} \times \frac{2n+1}{2^n*z^{2n}*z}.

b_n=\frac{(4n+2)*z^2}{2n+3}. Now take the limit as n \to \infty.

Ok so you've done your work right until the applying the limit. I get that the limit as n approaches infinity is 2z^2. Now you must find z such that |2z^2|&lt;1

 

[Mathman23]

Dear Jameson,

Thank You for checking and correcting my work :)

Is preferable to

|z^2|&lt;1/2 ?

Sincerely Yours
Fred

I'm going to do the Ratio Test as I type to check your work.

Calling b_n=\frac{a_{n+1}}{a_n} I get that b_n=\frac{(2^{n+1})(z^{2n+3})}{2n+3} \times \frac{2n+1}{(2^n)(z^{2n+1})}.

Now for simplification.

b_n=\frac{2*2^n*z^{2n}*z^3}{2n+3} \times \frac{2n+1}{2^n*z^{2n}*z}.

b_n=\frac{(4n+2)*z^2}{2n+3}. Now take the limit as n \to \infty.

Ok so you've done your work right until the applying the limit. I get that the limit as n approaches infinity is 2z^2. Now you must find z such that |2z^2|&lt;1

 

[HallsofIvy]

Hi

I'm suppose to show that the powerseries

\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1}

converge for all z \in mathbb{C}

By using the Ratio test:

I get

\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1} = \frac{2(2n+1)}{2n+3} |z^2| = \frac{2}{3}|z^2|, n \rightarrow \infty

I guess that can be re-written as |z^2| < 3/2 ?

But does that mean the ratio test fails?

Sincerley

Fred

First do not write
\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1} = \frac{2(2n+1)}{2n+3} |z^2|
The infinite sum is not equal to the ratio!

Second, are you asked to show that the series converges for all z or to find the radius of convergence?

I don't know what you mean by "Is preferable to |z|^2&lt; \frac{1}{2}?". As Jameson showed, the correct ratio is |2z^2|&lt; 1. Since 2 is positive, that is the same as 2|z^2|&lt; 1 or
|z|^2&lt; \frac{1}{2}. Finally, since |z| is a positive real number, |z|&lt; \frac{1}{\sqrt{2}}.

No, the ratio test does not "fail" but it does show that this series does not converge for all z.

 

[Mathman23]

Hello Hall,

I suppose to show that the series converge for all z. Since using ratio test shows that the series converge for |z| &lt; \frac{1}{\sqrt(2)}, then I need to use the comparison test?

Sincerely Yours

Fred

First do not write
\sum_{n=0} ^{\infty} \frac{2^n}{2n+1} z^{2n+1} = \frac{2(2n+1)}{2n+3} |z^2|
The infinite sum is not equal to the ratio!

Second, are you asked to show that the series converges for all z or to find the radius of convergence?

I don't know what you mean by "Is preferable to |z|^2&lt; \frac{1}{2}?". As Jameson showed, the correct ratio is |2z^2|&lt; 1. Since 2 is positive, that is the same as 2|z^2|&lt; 1 or
|z|^2&lt; \frac{1}{2}. Finally, since |z| is a positive real number, |z|&lt; \frac{1}{\sqrt{2}}.

No, the ratio test does not "fail" but it does show that this series does not converge for all z.

 

[HallsofIvy]

No! What the ratio test shows is that this series does not converge for all z! It has radius of convergence \frac{1}{\sqrt{2}}.

If you are not convinced, take z= 1. Does
\sum_{n=0} ^{\infty} \frac{2^n}{2n+1}
converge?

Does the sequence
\frac{2^n}{2n+1}
even converge to 0?

 

[Mathman23]

Thanks now I get ;)

Yours
Fred

No! What the ratio test shows is that this series does not converge for all z! It has radius of convergence \frac{1}{\sqrt{2}}.

If you are not convinced, take z= 1. Does
\sum_{n=0} ^{\infty} \frac{2^n}{2n+1}
converge?

 

[Mathman23]

No! What the ratio test shows is that this series does not converge for all z! It has radius of convergence \frac{1}{\sqrt{2}}.

If you are not convinced, take z= 1. Does
\sum_{n=0} ^{\infty} \frac{2^n}{2n+1}
converge?

Does the sequence
\frac{2^n}{2n+1}
even converge to 0?

Hello I discovered that I made an error in the post,

The original question should have looked like this:


Given a powers series

\sum_{n=0} ^\infty \frac{2^n}{(2n+1)!} z^{2n+1}

Show that the series converge for all z \in \mathbb{C}

By the ratio test,

I get that

\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n! \cdot z^{2n+3}}{(2n+3)!} \right| =\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n!}{(2n+3)!} \right| \left| z^{3} \right|

Since n \rightarrow \infty then

\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n! \cdot z^{2n+3}}{(2n+3)!} \right| \rightarrow 0.

Therefore it converges for all z?

Sincerely
Fred

 

[Curious3141]

Hello I discovered that I made an error in the post,

The original question should have looked like this:


Given a powers series

\sum_{n=0} ^\infty \frac{2^n}{(2n+1)!} z^{2n+1}

Show that the series converge for all z \in \mathbb{C}

By the ratio test,

I get that

\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n! \cdot z^{2n+3}}{(2n+3)!} \right| =\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n!}{(2n+3)!} \right| \left| z^{3} \right|

Since n \rightarrow \infty then

\mathop {\lim }\limits_{n \to \infty } \left| \frac{4(n+1) \cdot n! \cdot z^{2n+3}}{(2n+3)!} \right| \rightarrow 0.

Therefore it converges for all z?

Sincerely
Fred

I'm sorry, but I really cannot understand the algebra you're doing.

How did the '4', the '(n+1)!' etc. come into it? How did you get a z^3 term in the ratio?

First of all, please note that (2n+1)! \neq (2n+1)!

For a definition of the double factorial, please see your earlier thread entitled Powersums and ODEs and my exchange with Mathman. You're dealing with the same series as in that thread, and that is a double factorial.

What I get is like this. T_k represents the kth term in the series.

r = \frac{T_{n+1}}{T_n} = \frac{2z^2}{2n+3}

lim_{n \rightarrow \infty} r = lim_{n \rightarrow \infty} \frac{2z^2}{2n} = lim_{n \rightarrow \infty} \frac{z^2}{n} = 0

so the series converges by the ratio test.