Question 2.38: Investigate the convergence of the sequences (a) un = i^n/n ...

(a) The first few terms of the sequence are i, \frac{i^2}{2}, \frac{i^3}{3}, \frac{i^4}{4}, \frac{i^5}{5}, etc., or i,-\frac{1}{2}, \frac{-i}{3}, \frac{1}{4}, \frac{i}{5}, \ldots On plotting the corresponding points in the z plane, we suspect that the limit is zero. To prove this, we must show that

\left|u_n-l\right|=\left|i^n / n-0\right|<\epsilon \quad \text { when } n>N   (1)

Now

Let us choose N=1 / \epsilon. Then we see that (1) is true, and so the sequence converges to zero.
(b) Consider

For all n \geq 10 (for example), we have n \sqrt{2} /(n+1)>6 / 5=1.2. Thus \left|u_{n+1}\right|>1.2\left|u_n\right| for n>10, i.e., \left|u_{11}\right|>1.2\left|u_{10}\right|,\left|u_{12}\right|>1.2\left|u_{11}\right|>(1.2)^2\left|u_{10}\right|, and in general \left|u_n\right|>(1.2)^{n-10}\left|u_{10}\right|. It follows that \left|u_n\right| can be made larger than any preassigned positive number (no matter how large) and thus the limit of \left|u_n\right| cannot exist, and consequently the limit of u_n cannot exist. Thus, the sequence diverges.