Newton's Binomial summed to infinity

Hi All,

Just a quick question about Newton's Binomial.

When the upper bound of the summation is given the binomial is as follows:

$\displaystyle \sum_{i=0}^{n}{n \choose i}x^{n-i}y^{i}=(x+y)^{n}$

But what about when the upper bound is infinite? Does the result stay the same?

So is the following correct?

$\displaystyle \sum_{i=0}^{\infty}{n \choose i}x^{n-i}y^{i}=(x+y)^{n}$

Kind regards,

Chris

Re: Newton's Binomial summed to infinity

Quote:

Originally Posted by

**Vulpes** When the upper bound of the summation is given the binomial is as follows:

$\displaystyle \sum_{i=0}^{n}{n \choose i}x^{n-i}y^{i}=(x+y)^{n}$

But what about when the upper bound is infinite? Does the result stay the same?

So is the following correct? $\displaystyle \sum_{i=0}^{\infty}{n \choose i}x^{n-i}y^{i}=(x+y)^{n}$

@Vulpes, Do you not realize that $\displaystyle \infty$ **is not a number**?

If not, then you are not mathematically literate.

If you do understand that, then you know you question is meaningless.

Re: Newton's Binomial summed to infinity

Hi Vulpes,

I think your question depends on using the "generalized" binomial coefficient. That is for any real n and integer i,

$\displaystyle {n\choose i}={n(n-1)\cdots n-i+1\over i!}$

So if n is a positive integer and i > n, the binomial coefficient is 0. Thus the infinite series is just the finite sum. If n is a non-integer real, your infinite series is Newton's binomial series. See the article Binomial theorem - Wikipedia, the free encyclopedia

Re: Newton's Binomial summed to infinity

Quote:

Originally Posted by

**Vulpes** Hi All,

Just a quick question about Newton's Binomial.

When the upper bound of the summation is given the binomial is as follows:

$\displaystyle \sum_{i=0}^{n}{n \choose i}x^{n-i}y^{i}=(x+y)^{n}$

But what about when the upper bound is infinite? Does the result stay the same?

So is the following correct?

$\displaystyle \sum_{i=0}^{\infty}{n \choose i}x^{n-i}y^{i}=(x+y)^{n}$

Kind regards,

Chris

By definition$\displaystyle {n \choose i}=0$ when$\displaystyle i>n$, so yes it is correct as it is the same as the finite sum.

Re: Newton's Binomial summed to infinity

Quote:

Originally Posted by

**Plato** @Vulpes, Do you not realize that $\displaystyle \infty$ **is not a number**?

If not, then you are not mathematically literate.

If you do understand that, then you know you question is meaningless.

That logic that would make the sums of all infinite series meaningless!

.

Re: Newton's Binomial summed to infinity

Thank you johng and zephod,

It makes total sense that $\displaystyle {n \choose i}=0$ for $\displaystyle i > n$. Your explanation greatly helped me to understand the problem!

Plato, I do not quite understand why the question should be meaningless. In probability problems (which is what I'm working on) there are often experiments that can continue 'ad infinitum', which results in such an infinite sum of binomials.

Kind regards,

Chris

Re: Newton's Binomial summed to infinity

Plato was assuming that, since the upper bound was n, when you said "the upper bound is infinite", you meant n going to infinity. That makes your question meaningless- and Zephod's response about "i> n" irrelevant. Yes, there exist infinite sums that converge but those require that the summand go to 0, which would not happen here since binomial coefficients are always integers.

Re: Newton's Binomial summed to infinity

Quote:

Originally Posted by

**HallsofIvy** Plato was assuming that, since the upper bound was n, when you said "the upper bound is infinite", you meant n going to infinity. That makes your question meaningless- and Zephod's response about "i> n" irrelevant. Yes, there exist infinite sums that converge but those require that the summand go to 0, which would not happen here since binomial coefficients are always integers.

In the original question n is fixed (does not go to infinity), it is the range of summation for i that goes to infinity.

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