a belong to real set. Find the range of x such that the series converge: $\sum_{0}^{\infty}(\frac{a(a-1)...(a-k+1)}{k!})x^{k}$
Okay, i found that for |x| < 1 the series converge, by the ratio rest. Otherwise, and for |x| different of 1, the series diverges. But i think i am seeing a contradiction here:
See that the series is: $1 + ax + a(a-1)x²/2 + ...$.
I notice that $1 + ax + a(a-1)x²/2 + ... < 1 + ax + (a²)x²/2 + (a³)x³/3!...$
So that we can call the right side of the inequality as $e^{ax}$ And this is just a number! So that, technically, the series converge for any x. What am i missing here?
$$\sum_{k=0}^{\infty}\left(\frac{a(a-1)...(a-k+1)}{k!}\right)x^{k}=\sum _{k=0}^{\infty } \frac{a! x^k}{k! (a-k)!}=\sum _{k=0}^{\infty } x^k \binom{a}{k}$$ if $|x|<1$ the series converges to $(1+x)^a$