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

Question

Quizplus · Accepted Answer

This series is a telescoping series. When you write out the first few terms and the last few terms, you'll notice that most terms cancel out, leaving only the first few and the last few terms. Specifically, the first term is $\frac{1}{\sqrt{2}}$ (when $n=1$), and the second term is $\frac{1}{\sqrt{3}}$ (when $n=2$). As $n$ approaches infinity, the terms involving $n$ in the denominator approach 0, leaving $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}$ as the sum of the series.

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

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