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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 \(r<1\)

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\(f_{n}(z)\) uniformly converges for \(|z|<1\), while \(f_{n}^{\prime}(z)\) does not for \(|z|<1\), but does uniformly converge for \(|z| \leq r\) when \(r<1\).
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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 3: Showing Non-Uniform Convergence of \(f_{n}^{\prime}(z)\) for \(|z|

The sequence \(f_{n}^{\prime}(z)\) does not converge uniformly on \(|z|<1\), because the limit function would be \(0\) for \(z\neq 1\), and does not exist for \(z = 1\). Thus, the sequence \(\{f_{n}^{\prime}(z)\}_{n=1}^{\infty}\) is not uniformly convergent on the disk \(|z|<1\).

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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