Why is $\sum_{k=0}^{\infty} {1/2 \choose k} = \sqrt{2}$?

Question

As the header states, I would like to know, why $\sum_{k=0}^{\infty} {1/2 \choose k} = \sqrt{2}$. I know that $\sum_{k=0}^{n} {n \choose k} = 2^n$, so this seems naturally. The proof for the 2nd statement was done by induction, hence I do not have a clue how I could expand this proof. Additionally I tried to take a look at the definition of ${n \choose k}$ if n is a complex number, but I do not think rewriting the sum in this way helps.

Any hints? Thanks a lot!

Answer 1

Some fine points which are left out in other answers.

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First of all the general binomial theorem says that if $x, n$ are real numbers with $|x|<1$ then $$(1+x)^n=\sum_{k=0}^{\infty}\binom {n} {k} x^k\tag{1}$$ where the binomial coefficient $\binom{n} {k} $ is defined via $$\binom{n} {0}=1,\,\binom{n}{k}=\frac{n(n-1)(n-2)\dots (n-k+1)}{k!}, \, k=1,2,\dots\tag{2}$$ When $n$ is a positive integer then the series at the right hand side of equation $(1) $ is a finite series as all terms with $k>n$ are zero and this case is the easier one to handle and routinely proved using induction. If $n$ is not a positive integer then the series in $(1)$ is an infinite series and the proof of binomial theorem for this case is conveniently left out in most introductory calculus texts.

The result can be proved using Taylor's theorem or using the Cauchy formula for multiplication of infinite series. These proofs typically require the condition $|x|<1$ and with some more effort it can be proved that the result is true for $x=\pm 1$ also for certain values of $n$.

For the current problem $x=1$ and then it is known that the result holds for all $n>-1$ and thus for the question at hand we have $n=1/2$ and the desired sum is $(1+1)^{1/2}=\sqrt{2}$.

Answer 2

Hint: try expanding the polynomials $p_n(x) = (1+x)^n$, where $n$ is an integer. Write out the Pascal triangle beside them. You can discover that the coefficient $c_{n,m}$ of $x^m$ is exactly $\left(\begin{align}m\\n\end{align}\right)$. Now try to generalize this to the case where $n=1/2$, and substitute $x=1$, you'll get the answer.

Answer 3

This is an application of the generalized binomial theorem$$(a+b)^n=a^n+na^{n-1}b+\frac {n(n-1)}{2!}a^{n-2}b^2+\cdots$$ Where $n$ doesn’t necessarily have to be an integer. If $n$ is an integer, then the binomial theorem turns into the regular binomial theorem you probably learned in eigth grade. Now set $a=b=1$ and $n=1/2$.