Determine Convergence or Divergence of the Alternating Series$$\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 3 ( n + 1 ) ^ { 3 / 2 } } { n ^ { 3 / 2 } + 1 }$$

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The Answer of Determine Convergence or Divergence of the Alternating Series$$\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 3 ( n + 1 ) ^ { 3 / 2 } } { n ^ { 3 / 2 } + 1 }$$

Determine Convergence or Divergence of the Alternating Series $\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 3 ( n + 1 ) ^ { 3 / 2 } } { n ^ { 3 / 2 } + 1 }$

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Determine Convergence or Divergence of the Alternating Series ∑n=1∞(−1)n+13(n+1)3/2n3/2+1\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 3 ( n + 1 ) ^ { 3 / 2 } } { n ^ { 3 / 2 } + 1 }n=1∑∞​(−1)n+1n3/2+13(n+1)3/2​

Definitions:

Determine Convergence or Divergence of the Alternating Series $\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 3 ( n + 1 ) ^ { 3 / 2 } } { n ^ { 3 / 2 } + 1 }$