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Problem 3

# Show that $$f_{n}(z)=\frac{z^{n}}{n}$$ converges uniformly for $$|z|<1 .$$ Show also that $$f_{n}^{\prime}(z)$$ does not converge uniformly for $$|z|<1$$ but it does converge uniformly for $$|z| \leq r$$ for $$r<1$$

Expert verified
$$f_{n}(z)$$ uniformly converges for $$|z|<1$$, while $$f_{n}^{\prime}(z)$$ does not for $$|z|<1$$, but does uniformly converge for $$|z| \leq r$$ when $$r<1$$.
See the step by step solution

## Step 1: Showing Uniform Convergence of $$f_{n}(z)$$

To show that $$f_{n}(z)$$ converges uniformly, we'll use the Weierstrass M-Test. For $$|z|<1$$, we have $$|f_{n}(z)|\leq |z|^n/n$$. As $$|z|<1$$, $$|z|^n$$ tends to zero as $$n$$ tends to infinity. Thus, the sequence $$\sum_{n} |f_{n}(z)|$$ converges. By the Weierstrass M-Test, $$f_{n}(z)$$ converges uniformly for $$|z|<1$$.

## Step 2: Calculating $$f_{n}^{\prime}(z)$$

By the power rule of differentiation, $$f_{n}^{\prime}(z)= z^{n-1}$$ for $$n>1$$ and $$0$$ for $$n = 1$$.

## Step 4: Showing Uniform Convergence of $$f_{n}^{\prime}(z)$$ for $$|z| \leq r$$ for $$r For \(|z|\leq r$$ and for $$r<1$$, the sequence $$f_{n}^{\prime}(z)$$ based on the Weierstrass M-Test converges uniformly. This is because $$|f_{n}^{\prime}(z)|=|z|^{n-1}$$ for $$|z|\leq r$$ which tends to zero as $$n$$ tends to infinity, implying that the sequence $$f_{n}^{\prime}(z)$$ converges uniformly for $$|z|\leq r$$ for $$r<1$$.

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