Determine if the series converges or diverges. If the series converges, find its sum.
-$$\sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { n ^ { 3 / 2 } } - \frac { 1 } { ( n + 1 ) ^ { 3 / 2 } } \right)$$
A) converges; $\frac { 1 } { 6 \sqrt { 6 } }$
B) converges; $\frac { 1 } { 3 \sqrt { 3 } }$
C) diverges
D) converges; 1

Question

Quizplus · Accepted Answer

This series is a telescoping series. Each term cancels out the subsequent term except for the very first and last terms. As $n$ approaches infinity, the term $\frac{1}{(n+1)^{3/2}}$ approaches 0, leaving only the first term of the series, which is $\frac{1}{1^{3/2}} = 1$. Therefore, the sum of the series converges to 1.

Determine If the Series Converges or Diverges ∑n=1∞(1n3/2−1(n+1)3/2)\sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { n ^ { 3 / 2 } } - \frac { 1 } { ( n + 1 ) ^ { 3 / 2 } } \right)n=1∑∞​(n3/21​−(n+1)3/21​)

Determine If the Series Converges or Diverges $\sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { n ^ { 3 / 2 } } - \frac { 1 } { ( n + 1 ) ^ { 3 / 2 } } \right)$