Question

Test series for convergence:

1) conjugate [(1/n^i)]
2) (3i+n) / ( n^3+n+1)
3) [1/(n^i)]^2
4) e^(i cosh n)

Answer 1

Answer on Question #49482 – Math – Complex Analysis

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

Solution. Let us check the vanishing condition:

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

Answer: the series diverges.

2) $\sum_{n=1}^{\infty} \frac{3i + n}{n^3 + n + 1}$ — 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 $0 \leq \frac{\sqrt{n^2 + 9}}{n^3 + n + 1} \leq \frac{\sqrt{10n^2}}{n^3} = \frac{\sqrt{10}}{n^2}$, but the series $\sum_{n=1}^{\infty}\frac{\sqrt{10}}{n^2}$ converges because the power of $n$ in denominator is greater than 1. So, $\sum_{n=1}^{\infty}\frac{3i + n}{n^3 + n + 1}$ 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:

$\lim_{n\to \infty}\left|\left(\frac{1}{n^i}\right)^2\right| = \lim_{n\to \infty}\left|\left(\frac{1}{e^{iLn(n)}}\right)^2\right| = \lim_{n\to \infty}\left|e^{-i2Ln(n)}\right| = \lim_{n\to \infty}\left|e^{-i2(\ln n + i2\pi k)}\right| = \lim_{n\to \infty}\left|e^{4\pi k}e^{-i2\ln n}\right| = e^{4\pi k}\neq 0$, where $k$ is an 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, consider $e^{i\cosh n}$, $n$ is an 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:

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

Answer: the series diverges.

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