# Interval of convergence and sum of power series

1. May 16, 2015

### nuuskur

1. The problem statement, all variables and given/known data
Consider $\sum\limits_{n=0}^{\infty} \frac{n+1}{(2n)!}(x+1)^{2n+1}$. Find the interval of convergence and sum of the power series.

2. Relevant equations

3. The attempt at a solution
According to the textbook: given the power series $\sum a_n(x-c)^n$ the radius of convergence $R:=\frac{1}{\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}}$. It doesn't say anything about a series $\sum a_n(x-c)^{f(n)}$ where $f(n)= 2n +1$, for example. Can I use the said formula to calculate $R$ in this case? $a_n =\frac{n+1}{(2n)!}$ If so then:
$$R:=\frac{1}{\lim\limits_{n\to\infty} \frac{(n+2)(2n)!}{(2n+2)!(n+1)}} = \frac{1}{0}???$$

2. May 16, 2015

### HallsofIvy

Staff Emeritus
So your text book expects you to know that $a^{2n+ 1}= (a)(a^2)^n[/tex]. Yes, that is still a power series. Second, instead of relying on memorized formulas, you need to think about where those formulas come from and what they mean. Here you have [itex]a_n= \frac{n+1}{(2n)!}(x+ 1)^{2n+ 1}$ and $a_{n+1}= \frac{n+ 2}{(2n+2)!}(x+ 1)^{2n+ 3}$
Using the ratio test (which is where your formula comes from), we look at $$\frac{a_{n+1}}{a_n}= \frac{n+2}{(2n+2)!}(x+ 1)^{2n+3}\frac{2n!}{n+1}\frac{1}{x+ 1)^{2n+ 1}}= \frac{n+ 2}{n+1}\frac{(2n)!}{(2n+2)!}\frac{(x+ 1)^{2n+ 3}}{ (x+ 1)^{2n+ 1}}$$

So what is the limit of $\frac{n+2}{n+1}$, the limit of $\frac{(2n)!}{(2n+ 2)!}$, and the limit of $\frac{(x+ 1)^{2n+3}}{(x+ 1)^{2n+1}}$ as x goes to infinity?

(As for your limit that goes to "1/0", it is helpful to use the fact that
$$\frac{1}{\lim_{n\to \infty} a_n}= \lim_{n\to\infty}\frac{1}{a_n}$$
Especially when "$a_n$" is itself a fraction which can be "inverted".
$$\frac{1}{\lim_{n\to\infty}\frac{(n+2)(2n)!}{(2n+2)!(n+ 1)}}= \lim_{n\to\infty}\frac{(2n+2)!(n+1)}{(2n)!(n+ 1)}$$

I presume you know that $\frac{(2n+ 2)!}{(2n)!}= \frac{(2n)!(2n+1)(2n+2)}{(2n)!}$.)

3. May 16, 2015

### pasmith

By the ratio test, the series converges absolutely when $$\lim_{n \to \infty} \frac{|a_{n+1}||(x - c)^{f(n+1)}|}{|a_n||(x - c)^{f(n)}|} < 1.$$ Here this yields $$\lim_{n \to \infty} \left . \frac{(n+2)|x + 1|^{2n+3}}{(2n + 2)!} \right/ \frac{(n+1)|x + 1|^{2n+1}}{(2n)!} = \lim_{n \to \infty} \frac{n+2}{(n+1)(2n+1)(2n+2)}|x+1|^2 = 0.$$ What does that tell you about the values of $|x + 1|$ for which the series converges absolutely?

Don't forget the next part: Find the sum of the power series.

4. May 16, 2015

### wabbit

To apply the formula you use, you can start by setting $y=(1+x)^2$ and analysing the series in y.

5. May 16, 2015

### nuuskur

In response to #3:
From what I see the limit is independent of x, the limit is 0 regardless. Does the series converge in all of $\mathbb{R}$?

#4 - I don't understand why you would use that substitution, there is $(x+1)^{2n+1}$ which is $((x+1)^2)^{n+\frac{1}{2}} = y^{n+\frac{1}{2}}$. Elaborate on your idea a little more, please.

6. May 16, 2015

### wabbit

Well, did you try working out the series in term of $y$ ? Note that the term $y^{1\over2}=x+1$ is a common factor to all terms.

7. May 16, 2015

### nuuskur

Oh do you mean $(x+1)\sum \frac{n+1}{(2n)!}(x+1)^{2n}$?

8. May 16, 2015

### wabbit

Yes but then do that substitution : the series in y is exactly in the form required to apply your theorem - and of course the original series converges for a value $x=x_0$ if and only if the new series converges for $y=y_0=(x_0+1)^2$ .

9. May 16, 2015

### nuuskur

I still get the odd result of $\frac{1}{0}$. I'm not exactly certain how these things work, but it looks like the series is convergent in all of $\mathbb{R}$

If substituting $y:=(x+1)^2$
$y\sum_{n=0}^\infty \frac{n+1}{(2n)!}y^n$. Assuming it's true that the series converges everywhere, then integrating I would get $y^2\sum_{n=0}^\infty \frac{1}{(2n)!}y^n$ which is ALMOST the thing I would want it to be. Problem is with the $(2n)!$ if it only were $n!$.

Is something like this allowed? Let $2n=: m$ then substituting: $y^2\sqrt{y}\sum_{m=0}^\infty \frac{1}{m!}y^m$ and that would be useable, but I'm not convinced this doesn't break some rules.

10. May 16, 2015

### wabbit

The $1\over0$ result does mean "infinite radius of convergence" here, or more rigorously "the series converges everywhere". (That is something you can easily see for yourself if you go back to your textbook and go through the proof of that statement, or you can prove it directly.)

But I'm afraid the substitution was only useful for the purpose of applying your convergence criterion. To find the sum, it's actually easier to use $z=1+x$.

You cannot do that "m=2n" substitution - well, you can, but the power of y is not m then and the sum is over even values only, no good.

What you can do is try some transformation to find a series of the form $\sum \frac{z^{2n}}{(2n)!}$ or something closely related - presumably you already know the sum of that last series.

11. May 16, 2015

### nuuskur

Hmm, this substitution would yield $z\sum_{n=0}^\infty \frac{n+1}{(2n)!}z^{2n}$. If there was no $n+1$ it would be the hyperbolic cosine. Trouble is I don't know how to deal with the $n+1$. No doubt there is an explanation how to deal with this in the textbook, but the text is so abstract, it's difficult to understand "just like that".

12. May 16, 2015

### wabbit

Yes that $n+1$ is troublesome, but you're on the right track.

The easiest thing now is probably to write $\frac{n+1}{(2n)!}=\frac{n}{(2n)!}+\frac{1}{(2n)!}$.

Last edited: May 16, 2015
13. May 16, 2015

### nuuskur

Putting the $z$ back inside for a moment.
$\frac{1}{2}\sum\frac{2(n+1)}{(2n)!}z^{2n+1}$. Integrating with respect to $z$. $\frac{z^2}{2}\sum \frac{1}{(2n)!}z^{2n}$ (pulling $z^2$ back out). This would be the integral of the sum.

Another question: The series does converge Absolutely, however when we analyze a finite convergence radius, we also check the end points. If the series absolutely converges Everywhere does it therefore also converge everywhere? (it feels logical, there are no critical points to consider)

14. May 16, 2015

### wabbit

Yes good move, just be careful not to drop a term somewhere when you finalize the calculation:)

Indeed there is no issue with end points, if a series converges absolutely it converges - this doesn't even have to do with radius of convergence or anything, it's a pointwise statement.

Last edited: May 16, 2015
15. May 16, 2015

### nuuskur

If $\int S(z) = \frac{z^2}{2}\cosh z$ Then $S(z) = z\cosh z + \frac{z^2}{2}\sinh z$ and substituting back for $z= x+1$
$S(x) = (x+1)\cosh (x+1) + \frac{(x+1)^2}{2}\sinh (x+1)$

Side note:

I think it would look better if I did the integration properly, in my earlier "quick n dirty" version I should have also included the constant of integration. This way I can avoid it.
$$\int_0^z\frac{1}{2}\sum_{n=0}^\infty\frac{2(n+1)}{(2n)!}t^{2n+1}\mathrm{d}t$$

Last edited: May 16, 2015
16. May 16, 2015

### wabbit

Right, and if you want to be complete you can add a mention of a theorem that lets you swap the sum and integral here, $\sum\int=\int\sum$ , either using the fact that this is always true for power series inside their radius of convergence, or just referring to the general absolute convergence rule.