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We set (as usual) $\displaystyle{x \choose k} := \frac{x \cdot (x-1) \cdots (x-k+1)}{k!}$ for $x\in \mathbb{C}$.

Now we can define a function

$\displaystyle f(x) := \sum\limits_{k=0}^\infty {x \choose k}$.

Does anybody know how this function is called (I need its name, so that I can get more information about it)? I believe, it should be well-known, but I don't know its name.

Note that if we defined $\displaystyle{x \choose k} := \frac{x^k}{k!}$ instead, we would simply get the $\exp$ function - so my function is probably be related to it.

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How sure are you that the sum even converges for non-integral $x$? The series for the exponential function converges because the $k$ factors in the denominator of each term eventually dominate the $x$ factors in the numerator -- but here we eventually get factors of $-(k+1-x)$ in the numerator, which grow in magnitude as $k$ does. So it is not obvious even that $\binom xk\to 0$. –  Henning Makholm Jan 22 '12 at 20:47
The series certainly doesn't converge for $x \le -1$. @aelguindy: you need the stronger result that the numerator is less than $|x|^k$ in absolute value, which is false. –  Qiaochu Yuan Jan 22 '12 at 20:52
You're right, comment removed. –  aelguindy Jan 22 '12 at 20:56
Nubok, are you familiar with the binomial theorem for non-integer exponents, in the form of an expansion of the expression $(1+a)^x$? I'd suggest you start there. –  Gerry Myerson Jan 22 '12 at 22:15
Note $f(0)=1$ (the empty product is multiplicative unity) and $$f(x+1)=1+\sum_{k=1}^\infty\binom{x+1}{k}=1+\sum_{k=0}^\infty\left[\binom{x}{k}‌​+\binom{x}{k+1}\right]=2f(x).$$ That's a good hint $f(x)=2^x$. –  anon Jan 23 '12 at 3:04

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