Question
Determine if the series below converges absolutely, converges conditionally, or diverges.
\[
\sum_{n=1}^{\infty} \frac{(-1)^{n} n^{3 / 4}}{8 n^{3}+2}
\]

Step 1 ：First, we can factor the denominator with a little give and take: \[8n^3 + 2 = 8n^3 + 4n^{3/2} + 2 - 4n^{3/2} = (2n^{3/2} + 1)^2 - (2n^{1/2})^2 = (2n^{3/2} + 2n^{1/2} + 1)(2n^{3/2} - 2n^{1/2} + 1).\]

Step 2 ：Then \[\sum_{n=1}^\infty \frac{(-1)^n n^{3/4}}{8n^3 + 2} = \sum_{n=1}^\infty \frac{(-1)^n n^{3/4}}{(2n^{3/2} + 2n^{1/2} + 1)(2n^{3/2} - 2n^{1/2} + 1)}\]

Step 3 ：= \frac{1}{4} \sum_{n = 1}^\infty (-1)^n \frac{(2n^{3/2} + 2n^{1/2} + 1) - (2n^{3/2} - 2n^{1/2} + 1)}{(2n^{3/2} + 2n^{1/2} + 1)(2n^{3/2} - 2n^{1/2} + 1)}

Step 4 ：= \frac 1 4 \sum_{n=1}^\infty (-1)^n \left( \frac{1}{2n^{3/2} - 2n^{1/2} + 1} - \frac{1}{2n^{3/2} + 2n^{1/2} + 1} \right)

Step 5 ：= \frac 1 4 \sum_{n=1}^\infty (-1)^n \left( \frac{1}{(n^{1/2}-1)^2 + 1} - \frac{1}{(n^{1/2}+1)^2 + 1} \right)

Step 6 ：= \frac{1}{4} \left[ \left( \frac{1}{0^2 + 1} - \frac{1}{2^2 + 1} \right) - \left( \frac{1}{1^2 + 1} - \frac{1}{3^2 + 1} \right) + \left( \frac{1}{2^2 + 1} - \frac{1}{4^2 + 1} \right) - \dotsb \right].

Step 7 ：Observe that the sum telescopes. From this we find that the answer is \[\frac 1 4 \left( \frac{1}{0^2 + 1} - \frac 1 {1^2 + 1} \right) = \boxed{-\frac 1 4}.\]