Binomial series

1. Feb 26, 2015

The binomial series $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + ...$ only converges for $|x| < 1$ right?
Is it true that writing $(1 + x)^n$ differently (i.e. $x^n (1 + \frac{1}{x})^n$) extends the validity of this series to include values of $x$ such that $|x| > 1$?

2. Feb 26, 2015

Svein

No. A binomial series is only defined for a given n, Whatever the value of x, there exists an N such that $nx>1000$ for n>N.
No.

3. Feb 27, 2015

micromass

Yes. Although the series might converge for more values than just $|x|<1$. Complete details are here: http://en.wikipedia.org/wiki/Binomial_series#Conditions_for_convergence

Yes, If $|x|>1$, then your trick can be used. If $|x|<1$, then your trick doesn't work, but the original series does of course. If $|x|=1$ then the situation is a bit annoying.

4. Feb 27, 2015

Interesting. I once watched a video in which someone proves that the sum of all natural numbers is $-\frac{1}{12}$, and in one of the steps, he substituted $x = 1$ into the binomial series. Apparently, anything is possible if you're Euler!

5. Feb 27, 2015

micromass

Yes, that is valid, but only under conditions. For the standard convergence of series, plugging in $x=1$ is forbidden. But there are other definitions where series do not converge how we they usually do. Under those definitions, you do get $-1/12$.

6. Feb 27, 2015

Is it possible to approximate $\sqrt{2}$ using the binomial series?

7. Feb 27, 2015

micromass

Yes, but not directly. What you can do (for example) is to use $x=-1/2$ to approximate $1/\sqrt{2} = \frac{\sqrt{2}}{2}$. And then you can easily find $\sqrt{2}$.

Edit: I guess you can even do it directly, but convergence won't be very rapid.

Last edited: Feb 27, 2015

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