Vtyjkyuk

I want to find the order of the entire function $f(z)=sin(z)$. I have this result

Let
$$f(z)=sum_n=0^infty a_n z^n$$
be an entire function, non-constant and with finite order. Then, the order is given by
$$lambda=limsup_nto infty fracnlog n-log.$$

I've tried to apply this previous theorem to get (using Stirling's approximation)
$$lambda=limsup_nto infty fracnlog n-log=
limsup_nto infty fracnlog n(2n+1)log(2n+1)-2n-1=frac12.$$
However, the result should be $1$. I suspect that I cannot use this result because $a_2n=0$ is this case, and this will give problems in the expression.

How should I proceed?

First of all, the order of $sin(z)$ is the same as the order of $fracsin(z)z.$ This, in turn, can be written as

$$ fracsin(z)z = g(z^2),$$
for some entire $g$. Thus, the order of $sin(z)$ is twice the order of this $g$. But, from the Taylor coefficients of the sine function, we know that this $g$ can be written as
$$ g(z) = sum_n=0^infty b_n z^n,$$
where $b_n = frac(-1)^n(2n+1)!.$ Now you are allowed to use your result, together with Stirling's formula, to conclude.

P.S.: Notice also that the way you applied this result is slightly inaccurate, due to the fact that the given approximation holds only for $a_2n$, which implies there should be a $(2n)log(2n)$ in the numerator instead of $n log n.$