Answer to Question #299949 in Real Analysis for Dhruv rawat

Question

Test the series ∞ Σ n=1 (-1)^n-1. 3/(5n-2) for absolute and conditional convergence

Answer 1

ANSWER The series $\sum_{n=1}^{\infty}\frac{{3\cdot \left (-1 \right )}^{n-1}}{5n-2}$ is conditionally convergent .

EXPLANATION.

First we check absolute convergence . $\sum_{n=1}^{\infty}\ \left | \frac{{3\cdot \left (-1 \right )}^{n-1}}{5n-2} \right |=\sum_{n=1}^{\infty}\frac {3 }{5n-2}$ . Let $a _{n}=\frac{3}{5n-2}.$

$\lim_{n \rightarrow\infty}\frac{\frac{3}{5n-2}}{\frac{1}{n}}=\lim_{n \rightarrow\infty}\frac{3n}{5n-2}=\lim_{n \rightarrow\infty}\frac{3 }{5 -\frac{2}{n}}=\frac{3}{5}$ . Since $\lim_{n \rightarrow\infty}\frac {a_{n}} {\frac{1}{n}}=\frac{3}{5}>0$ and the series $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges ,then by the Limit Comparison Test $\sum_{n=1}^{\infty}\frac {3 }{5n-2}$ diverges. Therefore , the series does not converge absolutely.

The series $\sum_{n=1}^{\infty}\frac{{3\cdot \left (-1 \right )}^{n-1}}{5n-2}$ is alternating. Check the two conditions: $1) \lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\frac{3} {5n-2}=0$ . $2)a _{n+1} \leq a_{n}$ , because $5\cdot\left ( n+1 \right ) -2 \ge 5 n-2\\$ so

$\frac{3}{5\cdot\left ( n+1 \right ) -2}\le \frac{3}{5n-2}$ .

Since the two conditions of the Alternating Series Test are satisfied , $\sum_{n=1}^{\infty}\frac{{3\cdot \left (-1 \right )}^{n-1}}{5n-2}$ is conditionally convergent by the Alternating Series Test .