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Conclude if the given series converges or diverges and justify the conclusion

by RoRi

Test the following series for convergence or divergence. Justify the decision.

    \[ \sum_{n=1}^{\infty} \frac{n}{(4n-3)(4n-1)}. \]

Let

    \[ a_n = \frac{n}{(4n-3)(4n-1)}, \qquad b_n = \frac{1}{16n}. \]

Then we apply the limit comparison test,

    \begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\frac{n}{(4n-3)(4n-1)}}{\frac{1}{16n}} \\[9pt] &= \lim_{n \to \infty} \frac{16n^2}{(4n-3)(4n-1)} \\[9pt] &= \lim_{n \to \infty} \frac{16n^2}{16n^2 - 17n + 3} \\[9pt] &= \lim_{n \to \infty} \frac{16}{16 - \frac{17}{n} + \frac{3}{n^2}} \\[9pt] &= 1. \end{align*}

Hence, the series \sum a_n and \sum b_n either both converge or both diverge. But we know

    \[ \sum_{n=0}^{\infty} b_n = \frac{1}{16} \sum_{n=0}^{\infty} \frac{1}{n} \]

diverges (by the integral test, Example #1 on page 398 of Apostol). Hence,

    \[ \sum_{n=1}^{\infty} \frac{n}{(4n-3)(4n-1)} \]

diverges.

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