Question

Test the following series for convergence:
a) ∑(n goes from 1 to infinity) n^(1/n).
b)∑(n goes from 1 to infinity) (((-1)^n)sin(3n x))/n^3

Answer 1

Answer on Question #51428 - Math - Real Analysis

1) Test the following series for convergence:

(a) $\sum_{n=1}^{\infty} n^{1/n}$

Solution

Let us check the vanishing condition:

$\lim_{n \to \infty} n^{1/n} = \lim_{n \to \infty} e^{\frac{\ln n}{n}} = e^0 = 1 \neq 0$, Vanishing condition is the necessary condition for summability. So, the series $\sum_{n=1}^{\infty} n^{1/n}$ diverges.

Answer: the series diverges.

(b) $\sum_{n=1}^{\infty} \frac{(-1)^n \sin 3nx}{n^3}$

Solution

Let test it for absolute convergence:

$\sum_{n=1}^{\infty}\left|\frac{(-1)^n\sin 3nx}{n^3}\right| = \sum_{n=1}^{\infty}\frac{|\sin 3nx|}{n^3}$, apply the comparison test $0 \leq \frac{|\sin 3nx|}{n^3} \leq \frac{1}{n^3}$, but the series $\sum_{n=1}^{\infty}\frac{1}{n^3}$ converges because the power of $n$ in denominator is greater than 1. So, $\sum_{n=1}^{\infty}\frac{(-1)^n\sin 3nx}{n^3}$ is absolutely convergent for every real $x$.

Answer: the series is absolutely convergent for any $x \in \mathbb{R}$.

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