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

Question

Quizplus · Accepted Answer

This series is a telescoping series. When expanded, most terms cancel out, leaving only the first and last terms. The sum converges to $\frac { 1 } { \ln 3 }$.

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

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