Question

Test the series for convergence : Details

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

2) [n+i] / [4^n]

3) [ 1/ ( n^i ) ] ^2

4) e ^ ( i cosh n)

Answer 1

Answer on Question #49487 - Math - Complex Analysis

1) $\sum_{n=1}^{\infty} \frac{3i + n}{n^3 + n + 1}$ — test for convergence

Solution. Test it for absolute convergence:

$\sum_{n=1}^{\infty}\left|\frac{3i + n}{n^3 + n + 1}\right| = \sum_{n=1}^{\infty}\frac{\sqrt{n^2 + 9}}{n^3 + n + 1}$, use the comparison test $0 \leq \frac{\sqrt{n^2 + 9}}{n^3 + n + 1} \leq \frac{\sqrt{10n^2}}{3n^3} = \frac{\sqrt{10}}{3n^2}$, but the series $\sum_{n=1}^{\infty}\frac{\sqrt{10}}{3n^2}$ converges because the power of $n$ in denominator greater than 1. So, $\sum_{n=1}^{\infty}\frac{3i + n}{n^3 + n + 1}$ is absolutely convergent.

Answer: the series is absolutely convergent.

2) $\sum_{n=1}^{\infty} \frac{n + i}{4^n}$ — test for convergence

Solution. Test it for absolute convergence:

$\sum_{n=1}^{\infty}\left|\frac{n+i}{4^n}\right| = \sum_{n=1}^{\infty}\frac{\sqrt{n^2 + 1}}{4^n}$, use the ratio test

$\lim_{n\to \infty}\frac{\frac{\sqrt{(n + 1)^2 + 1}}{4^{n + i}}}{\frac{\sqrt{n^2 + 1}}{4^n}} = \lim_{n\to \infty}\frac{\sqrt{(1 + \frac{1}{n})^2 + \frac{1}{n^2}}}{4\sqrt{1 + \frac{1}{n^2}}} = \frac{1}{4} < 1$, that is why the series $\sum_{n = 1}^{\infty}\frac{\sqrt{n^2 + 1}}{4^n}$

converges. So, $\sum_{n=1}^{\infty} \frac{n+i}{4^n}$ is absolutely convergent.

Answer: the series is absolutely convergent.

3) $\sum_{n=1}^{\infty}\left(\frac{1}{n^i}\right)^2$ — test for convergence

Solution. Let us check the vanishing condition:

$\lim_{n\to \infty}\left|\left(\frac{1}{n^i}\right)^2\right| = \lim_{n\to \infty}\left|\left(\frac{1}{e^{iLn(n)}}\right)^2\right| = \lim_{n\to \infty}\left|e^{-i2Ln(n)}\right| = \lim_{n\to \infty}\left|e^{-i2(lnn + i2\pi k)}\right| = \lim_{n\to \infty}\left|e^{4\pi k}e^{-i2\ln n}\right| = e^{4\pi k}\neq 0$, where $k$ is a integer constant. Vanishing condition is the necessary condition for summability. So, the series $\sum_{n = 1}^{\infty}\left(\frac{1}{n^i}\right)^2$ diverges.

Answer: the series diverges.

4) $\sum_{n=1}^{\infty} e^{i \cosh n}$ — test for convergence

Solution. First of all, let's consider $e^{i \cosh n}$, $n$ is a integer number, then $\cosh n = \frac{e^n + e^{-n}}{2}$ is real number, this means that $\left|e^{i \cosh n}\right| = 1$

Let us check the vanishing condition:

$$\lim_{n \to \infty} \left| e^{i \cosh n} \right| = 1 \neq 0. \text{ Vanishing condition is the necessary condition for summability. So, the series } \sum_{n=1}^{\infty} e^{i \cosh n} \text{ diverges.}$$

Answer: the series diverges.

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