I Proving the Finite Binomial Series for k Non-Negative Integer

[Adrenaline101]

TL;DR Summary

What is the method to prove that a binomial series is not infinite when k is a non-negative integer.

Hello,

I was wondering how to prove that the Binomial Series is not infinite when k is a non-negative integer. I really don't understand how we can prove this. Do you have any examples that can show that there is a finite number when k is a non-negative integer?

Thank you!

 

[Gaussian97]

Do you mean the series expansion of $(1+x)^k$? Why should it be infinite?

 

[Adrenaline101]

Do you mean the series expansion of $(1+x)^k$? Why should it be infinite?

Yes, I mean that expansion. I was wondering how to prove that this series expansion is finite when k i a non-negative integer, because eventually, if the value of k is non-negative, this binomial series's values will eventually be 0.
I just cannot find a written proof that allows me to understand why this is happening.

 

[Gaussian97]

Shouldn't the fact that it is a finite degree polynomial be enough proof?

 

[PeroK]

Yes, I mean that expansion. I was wondering how to prove that this series expansion is finite when k i a non-negative integer, because eventually, if the value of k is non-negative, this binomial series's values will eventually be 0.
I just cannot find a written proof that allows me to understand why this is happening.

Any real analysis textbook should cover the convergence of power series. In general, a power series has a radius of convergence: e.g. for the binomial expansion:
$$(1 - x)^{-1} = 1 + x + x^2 + x^3 \dots$$ converges for $|x| < 1$ and diverges otherwise.

 

[PeroK]

Summary:: What is the method to prove that a binomial series is not infinite when k is a non-negative integer.

Hello,

I was wondering how to prove that the Binomial Series is not infinite when k is a non-negative integer. I really don't understand how we can prove this. Do you have any examples that can show that there is a finite number when k is a non-negative integer?

Thank you!

You mean $k$ is a negative integer, right? When $k$ is a positive integer, the expansion has a finite number of terms.

 

[Adrenaline101]

Oh I see... Sorry, I was stuck on a particular idea, which is more simple in reality.
Thank you!

 

[WWGD]

Let $n$ be a non-negative integer. If $n=0$ then for $a \neq 0, a^n=1$, If not, :

$(1 +x )^n = (1+x)(1+x)...(1+x)$ (n times) . The expression consists of a total of n expressions,each with two terms. If you want more rigor, do induction on n. If the expression $(1+x(^k$ is finite, so is $(1+x)^{k+1} =(1+x)^k (1+x)$ . If not, consider that the expanded expression consists of all monomials $a_n x^n$ given by $\Sigma_{i=0}^n nCi x^i$, where nCi is "n Choose i" . This is a finite sum with (n+1) terms.

 

[pasmith]

There's various way to do it:

(1) (1 + x)^k is a product of k polynomials of degree 1, and is therefore a polynomial of degree k.

(2) We have (1 + x)^\alpha = \sum_{n=0}^\infty a_n x^n where a_{n+1} = \frac{(\alpha - n)}{n+1}a_{n} with a_0 = 1. If \alpha is a non-negative integer then a_{\alpha + 1} = \frac{(\alpha - \alpha)}{\alpha + 1}a_\alpha = 0 and thus a_n = 0 for all n &gt; \alpha.

 

[Svein]

A compact way: <br /> 2^{n}=(1+1)^{n}=\int_{k=0}^{n}\begin{pmatrix}<br /> k\\<br /> n\\<br /> \end{pmatrix}1^{k}\cdot 1^{n-k}=\int_{k=0}^{n}\begin{pmatrix}<br /> k\\<br /> n\\<br /> \end{pmatrix}<br />