Interval of Convergence

Jun 2009
696
170
United States
I'm trying to determine the interval of convergence for the series:

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

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

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

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

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

Where am I going wrong?
 
Mar 2010
91
16
usa
The calculation should be:

\(\displaystyle \frac{u_{n+1}}{u_n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\cdot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}=\frac{(n+2)x}{(n+1)}\)
 
Dec 2009
1,506
434
Russia
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
 
Jun 2009
696
170
United States
The calculation should be:

\(\displaystyle \frac{u_{n+1}}{u_n}=\frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\cdot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}=\frac{(n+2)x}{(n+1)}\)
I know, but I don't see why that's true if \(\displaystyle (n+1)!=n!(n+1)\) I obtain:

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

So I cancel out the \(\displaystyle 2\cdot 3\cdot 4\cdot\cdot\cdot x^nn!\)

Notice that there are TWO \(\displaystyle (n+1)\) factors in the denominator. Where does that second factor go?
 
Jun 2009
696
170
United States
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 \(\displaystyle \frac{2\cdot 3\cdot 4\cdot\cdot\cdot(n+2)x^{n+1}}{(n+1)!}\frac{n!}{2\cdot 3\cdot 4\cdot\cdot\cdot(n+1)x^n}\).

See my last reply to elim
 

Defunkt

MHF Hall of Honor
Aug 2009
976
387
Israel
I know, but I don't see why that's true if \(\displaystyle (n+1)!=n!(n+1)\) I obtain:

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

So I cancel out the \(\displaystyle 2\cdot 3\cdot 4\cdot\cdot\cdot x^nn!\)

Notice that there are TWO \(\displaystyle (n+1)\) factors in the denominator. Where does that second factor go?
The first equality is incorrect. How did you get \(\displaystyle 2 \cdot 3 \cdot 4 \cdots n! \cdot x^n\) in the numerator? It should be \(\displaystyle 2 \cdot 3 \cdot 4 \cdots n \cdot x^n\) - you have an extra factorial there.
 
Jun 2009
696
170
United States
I think I've got it now.

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

That seems right.
 
Jun 2009
696
170
United States
The first equality is incorrect. How did you get \(\displaystyle 2 \cdot 3 \cdot 4 \cdots n! \cdot x^n\) in the numerator? It should be \(\displaystyle 2 \cdot 3 \cdot 4 \cdots n \cdot x^n\) - 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.

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

Defunkt

MHF Hall of Honor
Aug 2009
976
387
Israel
Oh, you're right; my bad. Looks like you just missed the \(\displaystyle n+1\) term from the numerator of the first fraction.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
The first thing I would do is simplify \(\displaystyle \frac{2\cdot 3\cdot 4\cdot\cdot\cdot (n+1)}{n!}\) to \(\displaystyle \frac{(n+1)!}{n!}= n+1\).

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