Is the Cumulant Generating Function correctly defined?

[donutmax]

Cumulative generating function is
$$K(t)=K_1(t)t+K_2(t)\frac{t^2}{2!}+K_3(t)\frac{t^3}{3!}+...$$
where
$$K_{n}(t)=K^{(n)}(t)$$

Now
$$K(t)=ln M(t)=ln E(e^{ty})=ln E(f(0)+f'(0)\frac {t}{1!}+f''(0)\frac{t^2}{2!}+...)=ln E(1+\frac{t}{1!}y+\frac{t^2}{2!} y^2+...)=ln [1+\frac{t}{1!} E(Y)+\frac{t^2}{2!} E(Y^2)+...]=ln [1+\frac{t}{1!}\mu'_1+\frac{t^2}{2!}\mu'_2+...]$$
where $$\mu'_n=E(Y^n)$$
$$=>K(0)=ln1=0$$

Also
$$K'(t)=\frac{1}{M(t)}M'(t)$$
where
$$M(0)=1; M'(t)=\mu'_1+\frac{t}{1}\mu'_2+\frac{t^2}{2!}\mu'_3+...$$
$$=>M'(0)=\mu'_1$$

In fact
$$M^{(n)}(0)=\mu'_n$$

So
$$K'(0)=\frac{\mu'_1}{1}=\mu'_1$$

Furthermore
$$K''(t)=\frac{M''(t)M(t)-[M'(t)]^2}{[M(t)]^2}$$
$$=>K''(0)=\frac{\mu'_2*1-(\mu'_1)^2}{1^2}=\mu'_2-(\mu'_1)^2=E(Y^2)-[E(Y)]^2=\sigma^2$$

Is this correct?

 

[donutmax]

Correction:
K(t) is:
$$K(t)=K_1t+K_2\frac{t^2}{2!}+...$$
where
$$K_n=K^{(n)}(0)$$