Interval of Convergence

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July 19th 2010, 10:25 AM
adkinsjr

Interval of Convergence

I'm trying to determine the interval of convergence for the series:

$\Sigma_{n=1}^{\infty}\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}{n!}$

$\lim_{n->\infty}\mid\frac{u_{n+1}}{u_n}\mid=\lim_{n->\infty}\mid\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\c dot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}\mid$

Using $(n+1)!=n!(n+1)$ , this simplifies:

$=\lim_{n-\infty}\mid\frac{(n+2)x}{(n+1)^2}\mid$

According to my solution manual, this is incorrect. The limit $\lim_{n->\infty}\mid\frac{u_{n+1}}{u_n}\mid=\lim_{n-\infty}\mid\frac{(n+2)x}{(n+1)}\mid$

Where am I going wrong?
July 19th 2010, 11:37 AM
elim

The calculation should be:

$\frac{u_{n+1}}{u_n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\c dot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}=\frac{(n+2)x}{(n+1)}$
July 19th 2010, 12:12 PM
Also sprach Zarathustra

lim |x(n+2)/(n+1)|=|x*lim (n+2)/(n+1)|=|x|*1

For the convergence you need:

|x|<1

Now check for x=1 and x=-1
July 19th 2010, 12:15 PM
adkinsjr

Quote:

Originally Posted by elim

The calculation should be:

$\frac{u_{n+1}}{u_n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\c dot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}=\frac{(n+2)x}{(n+1)}$

I know, but I don't see why that's true if $(n+1)!=n!(n+1)$ I obtain:

$\frac{2\cdot 3 \cdot 4\cdot\cdot\cdot (n+2)x^{n+1}}{n!(n+1)}\frac{n!}{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)x^n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}\frac{(n+2)x}{(n+1)(n+1)}=\frac{(n+2)x}{(n+1)^2 }$

So I cancel out the $2\cdot 3\cdot 4\cdot\cdot\cdot x^nn!$

Notice that there are TWO $(n+1)$ factors in the denominator. Where does that second factor go?
July 19th 2010, 12:18 PM
adkinsjr

Quote:

Originally Posted by Also sprach Zarathustra

lim |x(n+2)/(n+1)|=|x*lim (n+2)/(n+1)|=|x|*1

For the convergence you need:

|x|<1

Now check for x=1 and x=-1

I understand from that point, my problem is the simplification of $\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\c dot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}$ .

See my last reply to elim
July 19th 2010, 12:37 PM
Defunkt

Quote:

Originally Posted by adkinsjr

I know, but I don't see why that's true if $(n+1)!=n!(n+1)$ I obtain:

$\frac{2\cdot 3 \cdot 4\cdot\cdot\cdot (n+2)x^{n+1}}{n!(n+1)}\frac{n!}{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)x^n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}{2\cdot 3\cdot 4\cdot\cdot\cdot x^n n!}\frac{(n+2)x}{(n+1)(n+1)}=\frac{(n+2)x}{(n+1)^2 }$

So I cancel out the $2\cdot 3\cdot 4\cdot\cdot\cdot x^nn!$

Notice that there are TWO $(n+1)$ factors in the denominator. Where does that second factor go?

The first equality is incorrect. How did you get $2 \cdot 3 \cdot 4 \cdots n! \cdot x^n$ in the numerator? It should be $2 \cdot 3 \cdot 4 \cdots n \cdot x^n$ - you have an extra factorial there.
July 19th 2010, 12:41 PM
adkinsjr

I think I've got it now.

The sequence can be written as $\frac{(n+1)!x^n}{n!}$
$\frac{u_{n+1}}{u_n}=\frac{(n+1)!(n+2)x^{n+1}n!}{(n +1)!x^n(n+1)!}=\frac{(n+2)x}{n+1}$

That seems right.
July 19th 2010, 12:48 PM
adkinsjr

Quote:

Originally Posted by Defunkt

The first equality is incorrect. How did you get $2 \cdot 3 \cdot 4 \cdots n! \cdot x^n$ in the numerator? It should be $2 \cdot 3 \cdot 4 \cdots n \cdot x^n$ - you have an extra factorial there.

I don't see what you mean. The numerators are both the same (as far as I can tell), just rearranged.

$2\cdot 3\cdot 4\cdot (n+2)x^{n+1}n!=2\cdot 3\cdot 4\cdot n!x^n(n+2)x$
July 19th 2010, 12:52 PM
Defunkt

Oh, you're right; my bad. Looks like you just missed the $n+1$ term from the numerator of the first fraction.
July 19th 2010, 04:49 PM
HallsofIvy

The first thing I would do is simplify $\frac{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)}{n!}$ to $\frac{(n+1)!}{n!}= n+1$ .

The series you have is simply $\sum_{n=1}^\infty (n+1)x^n$ .
July 19th 2010, 04:55 PM
Also sprach Zarathustra

hmmm.... I don't know about this simplification...
July 19th 2010, 04:58 PM
Plato

Quote:

Originally Posted by Also sprach Zarathustra

hmmm.... I don't know about this simplification...

What do you not understand about that?
It is completely obvious.
July 19th 2010, 05:04 PM
Also sprach Zarathustra

Maybe it because of the night, I don't know! Anyway...

{2*4*6...*(n+1)}*{1*3*5*7*...*n} =(n+1)!

Edit: ooops(again!) I have not noticed that is not only the even numbers!

Sorry!
July 19th 2010, 07:07 PM
adkinsjr

Quote:

Originally Posted by HallsofIvy

The first thing I would do is simplify $\frac{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)}{n!}$ to $\frac{(n+1)!}{n!}= n+1$ .

The series you have is simply $\sum_{n=1}^\infty (n+1)x^n$ .

Yes, I realize I should have done that. I was foolishly jumping right into the ratio test, without looking for any simplifications in the series itself. I wasted too much time on this, lol.
July 20th 2010, 03:18 AM
HallsofIvy

Quote:

Originally Posted by Also sprach Zarathustra

Maybe it because of the night, I don't know! Anyway...

{2*4*6...*(n+1)}*{1*3*5*7*...*n} =(n+1)!

Edit: ooops(again!) I have not noticed that is not only the even numbers!

Sorry!

A product of even numbers can be reduced as:
$2(4)(6)\cdot\cdot\cdot(2n-2)(2n)= [(1)(2)][(2)(2)][(3)(2)]\cdot\cdot\cdot[(n-1)(2)][(n)(2)]$ $= n!2^n$ .

A product of odd numbers is more complicated- you have to put the even numbers in:
$(1)(3)(5)\cdot\cdot\cdot (2n-1)(2n+1)= \frac{(1)(2)(3)(4)(5)\cdot\cdot\cdot(2n-1)(2n)(2n+1)}{2(4)(6)\cdot\cdot\cdot(2n-2)(2n)}= \frac{(2n+1)!}{2^n n!}$ .

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