# Prove binomial series converges for |x|<1

Aziza

## Homework Statement

How do you prove that the binomial series (x+1)^p converges for |x|<1 ?

## The Attempt at a Solution

(x+1)^p = Ʃ$x^{n}\frac{p!}{(p-n)!n!}$
After doing ratio test I get |x|<1 . But now I have to test end points and this is my problem:

when x=1,
an = $\frac{p!}{(p-n)!n!}$

when x=-1,
an = (-1)^n$\frac{p!}{(p-n)!n!}$

Since the interval of convergence does not include these endpoints, I know these two series must diverge, but how to prove this? Is it because as n->∞, (p-n)! eventually becomes undefined, and so the limit of an as n->∞ is undefined, but in order for the series to converge, an must approach 0 as n->∞ ?

## Answers and Replies

Dickfore
But, your problem statement does not involve the endpoints you had mentioned (I'm assuming you consider $x \in \mathbb{R}$).

Aziza
But, your problem statement does not involve the endpoints you had mentioned (I'm assuming you consider $x \in \mathbb{R}$).

Well I used the ratio test, which is inconclusive when the limit as n->∞ of |$\frac{a_{n+1}}{a_{n}}$|=1.
ie, in this case, when |x|=1
So...I think you have to test the endpoints. Otherwise you cannot be certain that the interval of convergence contains them or not..

Dickfore
What is the actual statement of your problem?

Aziza
What is the actual statement of your problem?

"Use the ratio test to show that a binomial series converges for |x| < 1."

Dickfore
So, you did that.

Aziza
So, you did that.

no...this isn't the full proof...if I don't test the endpoints, it might be the case that the binomial series converges for |x|≤1 or -1<x≤1 or -1≤x<1, not for |x|<1

Dickfore
If it converges for $\vert x \vert \le 1$, then it definitely converges for $\vert x \vert < 1$. The converse is not true, however. Luckily, you are not required to prove the more stringent statement "the binomial series converges for $\vert x \vert \le 1$", so there is no need to test the endpoints.

Aziza
If it converges for $\vert x \vert \le 1$, then it definitely converges for $\vert x \vert < 1$. The converse is not true, however. Luckily, you are not required to prove the more stringent statement "the binomial series converges for $\vert x \vert \le 1$", so there is no need to test the endpoints.

ok yea I guess that is true...but...I still want to test the endpoints ! :)
so, how to do that in this case? how to prove that for x=1 and x=-1, the series diverge?

Dickfore
Definitely not by using the ratio test.

Aziza
Definitely not by using the ratio test.

yea i know that...
as i first wrote, i tried to prove that they diverge because their limit is undefined as n->∞. But, the limit of the general term of a series must tend to 0 as n->∞ in order for their to be a possibility of it converging. Thus, since the limit is undefined in this case, then it is not 0, and so the series must diverge. Is that the correct way of proving it?

Dickfore
yea i know that...
as i first wrote, i tried to prove that they diverge because their limit is undefined as n->∞. But, the limit of the general term of a series must tend to 0 as n->∞ in order for their to be a possibility of it converging. Thus, since the limit is undefined in this case, then it is not 0, and so the series must diverge. Is that the correct way of proving it?

No, since $a_{n} \rightarrow 0, \ n \rightarrow \infty$.

Aziza
No, since $a_{n} \rightarrow 0, \ n \rightarrow \infty$.

huh?

Dickfore
A more general definition of the binomial coefficient (valid for non-integer p) is:
$$\left(\begin{array}{c} p \\ n \end{array} \right) \equiv \frac{p (p - 1) \ldots (p - n + 1)}{n!}, \ n > 0,$$

So, you don't need to worry about the case $n - p < 0$ too much. If $p \in \mathbb{N}$, then, starting with $n_0 = p + 1$, the numerator of this fraction is always zero, so the series is a polynomial and is definitely convergent for any value of x. Suppose $p \notin \mathbb{N}$. Then, $\Gamma(-p)$ exists, and we may write:
$$\left(\begin{array}{c} p \\ n \end{array} \right) = \frac{(-1)^n \, (n - p - 1) \ldots (-p + 1) (-p)}{n!} = \frac{(-1)^n \, \Gamma(n - p)}{\Gamma(-p) \, \Gamma(n + 1)}$$
But, for large λ the Euler gamma function has the asymptotic form (Stirling's formula):
$$\Gamma(\lambda + 1) \sim \sqrt{2 \pi \lambda} \, \left( \frac{\lambda}{e} \right)^{\lambda}, \ \lambda \rightarrow \infty$$
and
$$a_n \sim \frac{(-1)^n}{\Gamma(-p)} \, \frac{(n - p - 1)^{n - p - \frac{1}{2}}}{n^{n + \frac{1}{2}}} \, e^{-n + p + 1 + n} \sim \frac{(-1)^n}{\Gamma(-p)} \, \left( \frac{e}{n} \right)^{p + 1} \rightarrow 0, \ n \rightarrow \infty$$
for $p > -1$.
For $p = -1$, the binomial coefficient is:
$$a_n = (-1)^n$$
and the series diverges on both ends.

For $p <-1$ (and non-natural number), the main coefficient diverges, and, therefore the series should diverge by the necessary condition for convergence.

You still haven't proven that the series converges or diverges at the endpoints for $p < -1$.

Aziza
ok now that i have looked at this more closely, i think that Ʃan converges at x=1 and x=-1...

an = $(\stackrel{p}{n})$$x^{n}$ = $(\stackrel{p}{n})$ when x=1

$(\stackrel{p}{n}) = \frac{p(p-1)(p-2)...(p-n+1)}{n!}$

but as n->∞, the numerator will eventually become 0 because n will become greater than p...in other words, $\left(\stackrel{p}{p+1}\right)$ = 0 for all p

the argument is same for x=-1

So i think what I was doing wrong was defining $(\stackrel{p}{n})$ to be $\frac{p!}{(p-n)!n!}$ , but that only works for n≤p, because when n>p, the denominator will contain the factorial of a negative number, which is not defined from what i understand...

so what confuses me now is why my book gives as a definition that the binomial series converges for |x|<1 and not for |x|≤1...unless my argument is wrong somehow?

Aziza
A more general definition of the binomial coefficient (valid for non-integer p) is:
$$\left(\begin{array}{c} p \\ n \end{array} \right) \equiv \frac{p (p - 1) \ldots (p - n + 1)}{n!}, \ n > 0,$$

So, you don't need to worry about the case $n - p < 0$ too much. If $p \in \mathbb{N}$, then, starting with $n_0 = p + 1$, the numerator of this fraction is always zero, so the series is a polynomial and is definitely convergent for any value of x. Suppose $p \notin \mathbb{N}$. Then, $\Gamma(-p)$ exists, and we may write:
$$\left(\begin{array}{c} p \\ n \end{array} \right) = \frac{(-1)^n \, (n - p - 1) \ldots (-p + 1) (-p)}{n!} = \frac{(-1)^n \, \Gamma(n - p)}{\Gamma(-p) \, \Gamma(n + 1)}$$
But, for large λ the Euler gamma function has the asymptotic form (Stirling's formula):
$$\Gamma(\lambda + 1) \sim \sqrt{2 \pi \lambda} \, \left( \frac{\lambda}{e} \right)^{\lambda}, \ \lambda \rightarrow \infty$$
and
$$a_n \sim \frac{(-1)^n}{\Gamma(-p)} \, \frac{(n - p - 1)^{n - p - \frac{1}{2}}}{n^{n + \frac{1}{2}}} \, e^{-n + p + 1 + n} \sim \frac{(-1)^n}{\Gamma(-p)} \, \left( \frac{e}{n} \right)^{p + 1} \rightarrow 0, \ n \rightarrow \infty$$
for $p > -1$.
For $p = -1$, the binomial coefficient is:
$$a_n = (-1)^n$$
and the series diverges on both ends.

For $p <-1$ (and non-natural number), the main coefficient diverges, and, therefore the series should diverge by the necessary condition for convergence.

You still haven't proven that the series converges or diverges at the endpoints for $p < -1$.

ohhh ok i will try to digest this now...ignore my last post then bc i posted not knowing you already did...

Aziza
ohhhh ok i see! so basically, the series converges at x=1 and x=-1 but only when p is in the natural numbers, but the point is to make it converge for all p...thank you!