∑n=1∞(−1)n3n2−24n5+n3−3\sum_{n=1}^{∞}{\left(-1\right)^n\frac{3n^2-2}{4n^5+n^3-3}}

asked by guest
on Feb 01, 2025 at 1:16 pm



You asked:

Evaluate the expression: ∑n=1∞(−1)n⋅3n2−24n5+n3−3\sum_{n = 1}^{\infty} {{\left( -1 \right)}^{n} \cdot \frac{3 {n}^{2} - 2}{4 {n}^{5} + {n}^{3} - 3}}

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