Is sequence \(z_{n}=e^{-\frac{n\pi i}{4}}\) bounded? convergent? Find their limit points.
Solution
Sequence is bounded, since each element has modulus \(1\). It does not converge, since sequence repeats. \(2\pi =\frac{n\pi }{4}\), hence \(n=8\). So only \(8\) elements are unique. Each of these is limit point. These are roots of \(\sqrt [8]{1}\).
Is sequence \(z_{n}=\frac{\left ( 3+4i\right ) ^{n}}{n!}\) bounded? convergent? Find their limit points.
Solution\[ z_{n}=\frac{\left ( re^{i\theta _{0}}\right ) ^{n}}{n!}\] But \(r=5\) and \(\theta _{0}=\arctan \left ( \frac{4}{3}\right ) \). The above becomes\begin{align*} z_{n} & =\frac{5^{n}e^{in\theta _{0}}}{n!}\\ & =\frac{5^{n}}{n!}e^{in\theta _{0}} \end{align*}
Since modulus of \(e^{in\theta _{0}}=1\), then we just need to look at \(\frac{5^{n}}{n!}\) to see if it is bounded or not. \(\lim _{n\rightarrow \infty }\frac{5^{n}}{n!}=0\). So it is bounded. Since \(n^{th}\) term goes to zero as \(n\rightarrow \infty \) it converges. The terms are \(\frac{5^{n}}{n!}\left ( \cos n\theta _{0}+i\sin n\theta _{0}\right ) \). It converges to zero, since \(\lim _{n\rightarrow \infty }\frac{5^{n}}{n!}=0\).
If \(z_{1},z_{2},\cdots \) converges to \(L\), and \(\bar{z}_{1},\bar{z}_{2},\cdots \) converges to \(\bar{L}\), show that \(z_{1}+\bar{z}_{1},z_{2}+\bar{z}_{2},\cdots \) converges to \(L+\bar{L}\)
Solution
This problem seems to be based on the idea that if sequence is convergent to \(L\), then for any \(\varepsilon \) no matter how small we can find an \(n\), such that \(\left \vert z_{n}-L\right \vert <\varepsilon \). So let us pick\begin{align*} \left \vert z_{n}-L\right \vert & <\frac{1}{2}\varepsilon \\ \left \vert \bar{z}_{n}-\bar{L}\right \vert & <\frac{1}{2}\varepsilon \end{align*}
Where in the above, we did the same for the other sequence. Now by triangle inequality \(\left \vert A+B\right \vert \leq \left \vert A\right \vert +\left \vert B\right \vert \), where now we treat \(A\) as \(\left ( z_{n}-L\right ) \) and \(B\) as \(\left ( \bar{z}_{n}-\bar{L}\right ) \), we have\begin{align*} \left \vert \left ( z_{n}-L\right ) +\left ( \bar{z}_{n}-\bar{L}\right ) \right \vert & \leq \left \vert z_{n}-L\right \vert +\left \vert \bar{z}_{n}-\bar{L}\right \vert \\ \left \vert \left ( z_{n}+\bar{z}_{n}\right ) -\left ( L+\bar{L}\right ) \right \vert & <\frac{1}{2}\varepsilon +\frac{1}{2}\varepsilon \end{align*}
The above is \(\left \vert \left ( z_{n}+\bar{z}_{n}\right ) -\left ( L+\bar{L}\right ) \right \vert <\varepsilon \). But this is the definition of a limit. It says that \(\left ( z_{n}+\bar{z}_{n}\right ) \) has limit \(L+\bar{L}\), which is what we are asked to show.
Are the following series convergent or divergent? Give a reason. \(\sum _{n=0}^{\infty }\frac{i^{n}}{n^{2}-2i}\)
Solution
The numerator has modulus \(1\). So we just need to consider \(\sum _{n=0}^{\infty }\frac{1}{\left \vert n^{2}-2i\right \vert }\). Since \(\frac{1}{n^{2}}\) converges and since \(\left \vert n^{2}-2i\right \vert >n^{2}\,\) (vectors, Argand diagram), then \(\frac{1}{\left \vert n^{2}-2i\right \vert }<\frac{1}{n^{2}}\), therefore it converges. We could also use the ratio test, but this is simpler.
Are the following series convergent or divergent? Give a reason. \(\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}\)
Solution
Since terms are \(\frac{1}{n^{\alpha }}\) where \(\left \vert \alpha \right \vert <1\), since \(\alpha =\frac{1}{2}\,\) here. Then we know it is divergent. It series becomes convergent for \(\alpha >1\). To show this, we can try the ratio test. But this gives the limit of \(1\), so ratio test is inconclusive. Using the integral test is best here. (notice that only upper limit is needed in this test, no need to use lower limit). We can use the integral test because the terms \(\frac{1}{\sqrt{n}}\) are monotonically decreasing.\begin{align*} \lim _{N\rightarrow \infty }\int ^{N}\frac{1}{x^{\frac{1}{2}}}dx & =\lim _{N\rightarrow \infty }\left ( 2\sqrt{x}\right ) ^{N}\\ & =\lim _{N\rightarrow \infty }2\sqrt{N}\\ & =\infty \end{align*}
Hence diverges.
Are the following series convergent or divergent? Give a reason. \(\sum _{n=1}^{\infty }n^{2}\left ( \frac{i}{3}\right ) ^{n}\)
Solution
Trying ratio test gives\begin{align*} \lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n+1\right ) ^{2}}{n^{2}}\frac{\left ( \frac{i}{3}\right ) ^{n+1}}{\left ( \frac{i}{3}\right ) ^{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n+1\right ) ^{2}}{n^{2}}\right \vert \left \vert \frac{\left ( \frac{i}{3}\right ) ^{n+1}}{\left ( \frac{i}{3}\right ) ^{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( \frac{i}{3}\right ) ^{n+1}}{\left ( \frac{i}{3}\right ) ^{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{i^{n+1}3^{n}}{i^{n}3^{n+1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{i}{3}\right \vert \\ & =\frac{1}{3} \end{align*}
Since limit is smaller than \(1\), then converges.
Find center and radius of convergence of series \(\sum _{n=0}^{\infty }\left ( \frac{a}{b}\right ) ^{n}z^{n}\)
Solution
For these type of problem, always compare it to standard form \(\sum _{n=0}^{\infty }A_{n}\left ( z-z_{0}\right ) ^{n}\). Where \(z_{0}\) is the center of disk. So we see that here \(z_{0}\) is the origin. Now to find \(R\) (the radius of convergence), it is given by the inverse of \(L=\lim _{n\rightarrow \infty }\left \vert \frac{A_{n+1}}{A_{n}}\right \vert \). Therefore we start by finding \(L\)\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( \frac{a}{b}\right ) ^{n+1}}{\left ( \frac{a}{b}\right ) ^{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{a^{n+1}b^{n}}{a^{n}b^{+1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{a}{b}\right \vert \\ & =\left \vert \frac{a}{b}\right \vert \end{align*}
Hence \(R=\left \vert \frac{b}{a}\right \vert \)
Find center and radius of convergence of series \(\sum _{n=0}^{\infty }\left ( n-i\right ) ^{n}z^{n}\)
Solution
The center is \(z_{0}=0\) by comparing to \(\sum _{n=0}^{\infty }A_{n}\left ( z-z_{0}\right ) ^{n}\). To find \(L\)\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n-i\right ) ^{n+1}}{\left ( n-i\right ) ^{n}}\right \vert \\ & =1 \end{align*}
Hence \(R=1\).
Find center and radius of convergence of series \(\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}z^{n}\)
Solution
The center is \(z_{0}=0\) by comparing to \(\sum _{n=0}^{\infty }A_{n}\left ( z-z_{0}\right ) ^{n}\). To find \(L\)\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{\left ( -1\right ) ^{n+2}}{n+2}}{\frac{\left ( -1\right ) ^{n+1}}{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( -1\right ) ^{n+2}n}{\left ( -1\right ) ^{n}\left ( n+2\right ) }\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{n}{\left ( n+2\right ) }\right \vert \\ & =1 \end{align*}
Hence \(R=1\).
Find center and radius of convergence of series \(\sum _{n=1}^{\infty }\frac{4^{n}}{\left ( 1+i\right ) ^{n}}\left ( z-5\right ) ^{n}\)
Solution
The center is \(z_{0}=5\) by comparing to \(\sum _{n=0}^{\infty }A_{n}\left ( z-z_{0}\right ) ^{n}\). To find \(L\)\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{4^{n+1}}{\left ( 1+i\right ) ^{n+1}}}{\frac{4^{n}}{\left ( 1+i\right ) ^{n}}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{4^{n+1}\left ( 1+i\right ) ^{n}}{4^{n}\left ( 1+i\right ) ^{n+1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{4\left ( 1+i\right ) ^{n}}{\left ( 1+i\right ) ^{n+1}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{4}{1+i}\right \vert \\ & =\lim _{n\rightarrow \infty }\frac{4}{\left \vert 1+i\right \vert }\\ & =\lim _{n\rightarrow \infty }\frac{4}{\sqrt{2}} \end{align*}
Hence \[ R=\frac{\sqrt{2}}{4}\]
Find center and radius of convergence of series \(\sum _{n=1}^{\infty }\frac{\left ( 4n\right ) !}{2^{n}\left ( n!\right ) ^{4}}\left ( z+\pi i\right ) ^{n}\)
Solution
The center is \(z_{0}=-\pi i\) by comparing to \(\sum _{n=0}^{\infty }A_{n}\left ( z-z_{0}\right ) ^{n}\). To find \(L\)\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{\frac{\left ( 4(n+1)\right ) !}{2^{\left ( n+1\right ) }\left ( \left ( n+1\right ) !\right ) ^{4}}}{\frac{\left ( 4n\right ) !}{2^{n}\left ( n!\right ) ^{4}}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 4(n+1)\right ) !2^{n}\left ( n!\right ) ^{4}}{\left ( 4n\right ) !2^{\left ( n+1\right ) }\left ( \left ( n+1\right ) !\right ) ^{4}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 4(n+1)\right ) !\left ( n!\right ) ^{4}}{\left ( 4n\right ) !\left ( \left ( n+1\right ) !\right ) ^{4}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 4n+4\right ) !\left ( n!\right ) ^{4}}{\left ( 4n\right ) !\left ( \left ( n+1\right ) !\right ) ^{4}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 4n+4\right ) \left ( 4n+3\right ) \left ( 4n+2\right ) \left ( 4n+1\right ) \left ( 4n\right ) !\left ( n!\right ) ^{4}}{\left ( 4n\right ) !\left ( \left ( n+1\right ) !\right ) ^{4}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 4n+4\right ) \left ( 4n+3\right ) \left ( 4n+2\right ) \left ( 4n+1\right ) \left ( n!\right ) ^{4}}{\left ( \left ( n+1\right ) !\right ) ^{4}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 4n+4\right ) \left ( 4n+3\right ) \left ( 4n+2\right ) \left ( 4n+1\right ) \left ( n!\right ) ^{4}}{\left ( \left ( n+1\right ) n!\right ) ^{4}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\left ( 4n+4\right ) \left ( 4n+3\right ) \left ( 4n+2\right ) \left ( 4n+1\right ) }{\left ( n+1\right ) ^{4}}\right \vert \\ & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{\allowbreak 256n^{4}+640n^{3}+560n^{2}+200n+24}{n^{4}+4n^{3}+6n^{2}+4n+1}\right \vert \end{align*}
Hence\begin{align*} L & =\frac{1}{2}\lim _{n\rightarrow \infty }\left \vert \frac{256+640\frac{1}{n}+560\frac{1}{n^{2}}+200\frac{1}{n^{3}}+\frac{24}{n^{4}}}{1+4\frac{1}{n}+6\frac{1}{n^{2}}+4\frac{1}{n^{3}}+\frac{1}{n^{4}}}\right \vert \\ & =\frac{1}{2}\left ( 256\right ) \\ & =128 \end{align*}
Hence \[ R=\frac{1}{128}\]