For $p\in (0, 1)$ and $n\rightarrow \infty$, how do I evaluate a general term in the binomial expansion of $(p+1-p)^n$?

Question

This may be a slightly trivial question but I got stuck on it for a while and would be grateful if someone can point out the catch to me.

The binomial expansion of $(p+1-p)^n$ is $$\sum_{i=0}^n\binom{n}{i}p^i(1-p)^{n-i}$$ for $p\in(0,1)$ and $n\rightarrow\infty$, a general term in the binomial expansion, $\binom{n}{i}p^i(1-p)^{n-i}$, in my opinion will tend to zero. This is because the rate at which $\binom{n}{i}$ grows is slower than the rate at which $p^i(1-p)^{n-i}$ decays given that the former is an $i$-th order polynomial in $n$ whereas the latter is an exponential.

If this is true, then every term in the sum will be zero and the sum will be zero, too, which is certainly not the case since $$\lim_{n\rightarrow\infty}(p+1-p)^n=1$$

Many thanks for your time.

Answer 1

Each term will go to zero but you have more and more terms as $n \to \infty$.

Answer 2

I think not using the Binomial Theorem is making things very unnecessarily complicated.

From Binomial Theorem, we have

$${(a+b)^n = \sum_{i=0}^{n}{n\choose i}a^ib^{n-i}}$$

Now, take ${1^n}$ as you did. Now take some non-zero number $p$. Any non-zero number. It does not have to be between $0$ and $1$. Then it's always true that

$${1^n = (p + (1-p))^n}$$

And now using Binomial Theorem with ${a=p}$, ${b=(1-p)}$ you get

$${(p+(1-p))^n = \sum_{i=0}^{n}{n\choose i}p^i(1-p)^{n-i}}$$

But ${(p+(1-p))^n = 1^n = 1}$. So the sum always ${=1}$, always!

$${\sum_{i=0}^{n}{n\choose i}p^i(1-p)^{n-i}=1}$$

It does not matter if ${n}$ was ${1}$, ${2}$, ${10}$, ${500}$, ${1000^2}$... the sum is always $1$. And there is nothing strange about this. If you put larger and larger values of $n$ - I agree, each term will get smaller and smaller. However, you are adding more and more terms to compensate. To really hammer this home...

$${\frac{1}{2}=\frac{1}{2}}$$

But also

$${\frac{1}{4} + \frac{1}{4} = \frac{1}{2}}$$

But also

$${\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}}$$

But also

$${\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}=\frac{1}{2}}$$

... And so on. If I keep doing this, each term is becoming much, much smaller. But it always sums to ${\frac{1}{2}}$, by construction. There are more and more terms to compensate for the fact that each term is smaller. This also helps to motivate the fact that the expression

$${0\times \infty}$$

is an indeterminate form. We cannot assign any value to ${0\times \infty}$ - since the answer depends on context. So indeed for any real number ${p\neq 0}$, you have that

$${\lim_{n\to \infty}\sum_{i=0}^{n}{n\choose i}p^i(1-p)^{n-i}=1\neq 0}$$

and there isn't anything strange about this result.