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Determine wether the following function converges or diverges by comparison test: $\sum_{n=1}^\infty \frac{3n-1}{(n+1)^3}$

Upon inspection I can clearly see that the series converges. However I am unsure how to know which series I can compare the series above with the prove this. Can someone tell me which series they would use and how they decided upon that?

Thank you soo much!

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$$ \sum_{n=1}^\infty\frac{3n-1}{(n+1)^3}\le\sum_{n=1}^\infty\frac{3n}{(n+1)^3}\le\sum_{n=1}^\infty\frac{3n}{n^3}=3\sum_{n=1}^\infty\frac1{n^2}<\infty. $$

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$$\sum_{n\geq 1}\frac{3n-1}{(n+1)^3} = 3\sum_{n\geq 1}\frac{1}{(n+1)^2}-4\sum_{n\geq 1}\frac{1}{(n+1)^3} = 3(\zeta(2)-1)-4(\zeta(3)-1) = \color{red}{3\zeta(2)-4\zeta(3)+1}.$$

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