Question

Test series for convergence:
1) [ i^n/(2^(n+2)) ]
2) (n!)^2/e^n
3) 1/[root(i+n)]^n
4) e^(i coshn)
5) [n+i/(4^n)]

Answer 1

Answer on Question #49481 – Math – Complex Analysis

Test series for convergence:

1) [ i^n/(2^(n+2)) ]

2) (n!)^2/e^n

3) 1/[root(i+n)]^n

4) e^(i coshn)

5) [n+i/(4^n)]

Solution

The complex series $\sum_{n=0}^{\infty} c_n$ is said to converge absolutely if the real series $\sum_{n=0}^{\infty} |c_n|$ converges. The following statement can be proved: if a complex series converges absolutely, then it converges. We shall use this fact in the next problems.

1) $\left|\frac{i^n}{2^{n+2}}\right| = \frac{1}{2^{n+2}} = \frac{1}{4} \cdot \frac{1}{2^n}$ is a geometric sequence with common ratio $q = \frac{a_{n+1}}{a_n} = \frac{1}{2} < 1$, so the series converges.

2) $\frac{c_{n+1}}{c_n} = \frac{\left((n+1)\right)^2}{e^{n+1}} \cdot \frac{(n!)^2}{e^n} = \frac{(n+1)^2}{e} > 1$ for all $n \geq 1$, $\lim_{n \to \infty} \frac{c_{n+1}}{c_n} = \lim_{n \to \infty} \frac{(n+1)^2}{e} > 1$ (it is plus infinity), hence by d'Alambert's ratio test, the series diverges.

3) $\left|\frac{1}{(\sqrt{i+n})^n}\right| = \left|\frac{1}{(i+n)^2}\right| = \frac{1}{|i+n|^2} < \frac{1}{n^2}$. Note that $\sqrt[n]{\frac{1}{n}} = \frac{1}{\sqrt{n}} < 1$, $\lim_{n \to \infty} \sqrt{|a_n|} < \lim_{n \to \infty} \sqrt[n]{\frac{1}{n}} = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$ and by the Cauchy ratio test, the series converges.

4) $\left| e^{i \cosh(n)} \right| = 1 \nrightarrow 0$ (it does not tend to zero) as $n \to \infty$ (the necessary condition of convergence does not hold true in this case), so the series diverges.

5) $\left|\frac{n + i}{4^n}\right| < \frac{2n}{4^n}$. Note that $\lim_{n \to \infty} \sqrt[n]{\frac{2n}{4^n}} = \frac{1}{4} < 1$ and by the Cauchy ratio test, the series converges.

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