Cfrgtkky

I am a bit confused about how to approach this question / not sure if I'm on the right track.

I know we can write $e^z = sum_{n=0}^{infty} frac{z^n}{n!}$. Further:

$e^z$
$= e^{z_0}(e^{z-z_0})$
$= e^{z_0} sum_{n=0}^{infty} frac{(z - z_0)^n}{n!}$

Replacing $z_0$ with $1$ we get:

$e^z= e^{1} sum_{n=0}^{infty} frac{(z - 1)^n}{n!}$

Is the above equation the correct Taylor Series for $e^z$ around $z_0 = 1$?

Also: if $|z-1| < R$ for some fixed $R > 0$ then:

$e^{1} sum_{n=0}^{infty} frac{|(z - 1)^n|}{n!} leq e sum_{n=0}^{infty} frac{|R^n|}{n!}$

Using the ratio test:

$L = lim_{n to infty}|frac{R^{n+1}}{(n+1)!}*frac{n!}{R^n}| = frac{R}{n+1}$

Since $L = 0$ for any fixed $R$, does this mean that the Taylor Series converges everywhere?

I'm a little confused on how to put everything together, so any insight is appreciated!

That's the correct series. You can also get it directly from the formula:

$$ f(z) = sum_{n = 0}^infty frac{f^{(n)}(1)}{n!}(z - 1)^n. $$

Both methods are good.

For the second question, you have
$$ e^{1} sum_{n=0}^{infty} frac{|(z - 1)^n|}{n!} leq e sum_{n=0}^{infty} frac{R^n}{n!} $$

and the ratio test shows that this converges. Thus the Taylor series converges absolutely for every $|z - 1| < R$, for every $R > 0$. Thus it converges for every $z in mathbb C$ (since you can always find an $R$ such that $|z - 1| < R$). This is perfectly valid reasoning.

A common argument here is exactly what you did, but phrased using the Weierstrass M-test. Namely, take

$$ f_n(z) = frac{e^1}{n!}(z - 1)^n text{ and } M_n = frac{e^1}{n!} R^n $$

Then on the set ${|z - 1| le R}$,

$$ |f_n(z)| le M_n $$

Since the sum $sum_{n ge 0} M_n$ converges, the M-test says that

$$ sum_{n = 0}^infty f_n(z) = sum_{n = 0}^infty frac{e^1}{n!}(z - 1)^n $$

converges absolutely and uniformly on the same set, ${|z - 1| le R}$. Absolute convergence you've seen, uniform convergence implies, for instance, that the resulting sum is continuous everywhere.