Answer:
Step-by-step explanation:
Given a series [tex]\sum_{k=1}^\infty f(z)[/tex], the Weierstrass M-test tell us that if we find a sequence of positive numbers [tex]M_n[/tex] such that [tex]|f(z)|\leq M_n[/tex] in a certain domain D, and the series [tex]\sum_{n=1}^\infty M_k[/tex] converges, then the series [tex]\sum_{k=1}^\infty f_k(z)[/tex] converges uniformly in the domain D.
So, our objective is to find the so called sequence [tex]M_k[/tex]. The main idea is to bound the sequence of functions [tex]\frac{z^k}{z^k + 1}[/tex].
Now, notice that the values of z are always positive, so [tex]z^k[/tex] is always positive, so [tex]z^k+1\geq 1[/tex] for all values of z in [tex]\overline{D}[/tex]. Then,
[tex]\Big| \frac{z^k}{z^k + 1}\Big| \leq z^k,[/tex]
because if we make the values of the denominator smaller, the whole fraction becomes larger.
Moreover, as z is in the interval [0,r], we have that [tex]z\leq r[/tex] and as consequence [tex]|z^k|\leq r^k[/tex]. With this in addition to the previous bound we obtain
[tex]\Big| \frac{z^k}{z^k + 1}\Big| \leq |z^k|\leq r^k.[/tex]
With this, our sequence is [tex]M_k = r^k[/tex] and the corresponding series is [tex]\sum_{k=1}^\infty r^k[/tex], which is a geometric series with ratio less than 1, hence it is convergent.
Then, as consequence of Weierstrass M-test we have the uniform convergence of the series in the given domain.
