Question

Test the convergence of the series : 2/1^3 - 2/2^3 + 3/3^3 + 5/4^3 ........

Answer 1

Answer on Question #56316 – Math – Real Analysis

Test the convergence of the series: $\frac{2}{1^3} - \frac{2}{2^3} + \frac{3}{3^3} + \frac{5}{4^3} + \cdots$

Solution

If $a_1 = \frac{2}{1^3}$, $a_2 = -\frac{2}{2^3}$, $a_3 = \frac{3}{3^3}$, $a_4 = \frac{5}{4^3}$ are terms of the given series, then we can rewrite $a_n = \frac{x_n}{n^3}$, where (for example) $|x_n| \leq 3n$ for all $n = 1, 2, \ldots$. Because the series

$$\sum_{n=1}^{\infty} \frac{3n}{n^3} = 3 \sum_{n=1}^{\infty} \frac{1}{n^2}$$

is convergent, the series

$$\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \frac{|x_n|}{n^3} \leq 3 \sum_{n=1}^{\infty} \frac{1}{n^2}$$

is also convergent by the Comparison Test. Thus, the series is absolutely convergent.

Hence, the series $\sum_{n=1}^{\infty} a_n$ is convergent

**Answer**: the given series converges.

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