# Interval of Convergence

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?

#### elim

The calculation should be:

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

#### 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

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?

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
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.

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.

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
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
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 $$\displaystyle \sum_{n=1}^\infty (n+1)x^n$$.