Teloscoping Sories In Exercises 39-44, find a formula for the $$ n $$th partial sum of the series and use it to determine if the series converges or diverges. If a series coaverges, find its sum. $$ \begin{array}{ll} ext { 39. } \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1} ight) & ext { 40. } \sum_{n=1}^{\infty}\left(\frac{6}{n^{2}}-\frac{6}{(n+1)^{2}} ight) \ ext { 41. } \sum_{n=1}^{\infty}(\ln \sqrt{n+1}-\operatorname{la} \sqrt{n}) & ext { 42. } \sum_{n=1}^{\infty}( an (n)- an (n-1)) \ ext { 43. } \sum_{n=1}^{\infty}\left(\operatorname{lin}^{n-1}\left(\frac{1}{n+1} ight)-\sin ^{-1}\left(\frac{1}{n+2} ight) ight)\end{array} $$

Question

Teloscoping Sories In Exercises 39-44, find a formula for the $$ n $$th partial sum of the series and use it to determine if the series converges or diverges. If a series coaverges, find its sum. $$ \begin{array}{ll}	ext { 39. } \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}ight) & 	ext { 40. } \sum_{n=1}^{\infty}\left(\frac{6}{n^{2}}-\frac{6}{(n+1)^{2}}ight) \ 	ext { 41. } \sum_{n=1}^{\infty}(\ln \sqrt{n+1}-\operatorname{la} \sqrt{n}) & 	ext { 42. } \sum_{n=1}^{\infty}(	an (n)-	an (n-1)) \ 	ext { 43. } \sum_{n=1}^{\infty}\left(\operatorname{lin}^{n-1}\left(\frac{1}{n+1}ight)-\sin ^{-1}\left(\frac{1}{n+2}ight)ight)\end{array} $$

UpStudy AI · Accepted Answer

To solve the problem, we will analyze each series step by step, finding the $$ n $$th partial sum and determining whether the series converges or diverges.

### 39. $$ \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) $$

1. **Finding the $$ n $$th partial sum $$ S_n $$**:
   $$
   S_n = \sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)
   $$
   This is a telescoping series. When we expand it, we get:
   $$
   S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right)
   $$
   Most terms cancel out, leaving:
   $$
   S_n = 1 - \frac{1}{n+1}
   $$

2. **Finding the limit as $$ n \to \infty $$**:
   $$
   \lim_{n \to \infty} S_n = 1 - 0 = 1
   $$
   Thus, the series converges and its sum is $$ 1 $$.

### 40. $$ \sum_{n=1}^{\infty}\left(\frac{6}{n^{2}}-\frac{6}{(n+1)^{2}}\right) $$

1. **Finding the $$ n $$th partial sum $$ S_n $$**:
   $$
   S_n = \sum_{k=1}^{n}\left(\frac{6}{k^{2}}-\frac{6}{(k+1)^{2}}\right)
   $$
   This is also a telescoping series:
   $$
   S_n = 6\left(\frac{1}{1^{2}} - \frac{1}{(n+1)^{2}}\right) = 6\left(1 - \frac{1}{(n+1)^{2}}\right)
   $$

2. **Finding the limit as $$ n \to \infty $$**:
   $$
   \lim_{n \to \infty} S_n = 6\left(1 - 0\right) = 6
   $$
   Thus, the series converges and its sum is $$ 6 $$.

### 41. $$ \sum_{n=1}^{\infty}(\ln \sqrt{n+1}-\ln \sqrt{n}) $$

1. **Finding the $$ n $$th partial sum $$ S_n $$**:
   $$
   S_n = \sum_{k=1}^{n}(\ln \sqrt{k+1}-\ln \sqrt{k}) = \sum_{k=1}^{n}\frac{1}{2}(\ln(k+1) - \ln(k))
   $$
   This simplifies to:
   $$
   S_n = \frac{1}{2}(\ln(\sqrt{n+1}) - \ln(\sqrt{1})) = \frac{1}{2}\ln(n+1)
   $$

2. **Finding the limit as $$ n \to \infty $$**:
   $$
   \lim_{n \to \infty} S_n = \infty
   $$
   Thus, the series diverges.

### 42. $$ \sum_{n=1}^{\infty}(\tan(n)-\tan(n-1)) $$

1. **Finding the $$ n $$th partial sum $$ S_n $$**:
   $$
   S_n = \sum_{k=1}^{n}(\tan(k)-\tan(k-1))
   $$
   This is a telescoping series:
   $$
   S_n = \tan(n) - \tan(0) = \tan(n)
   $$

2. **Finding the limit as $$ n \to \infty $$**:
   $$
   \lim_{n \to \infty} S_n \text{ does not exist (as } \tan(n) \text{ oscillates).}
   $$
   Thus, the series diverges.

### 43. $$ \sum_{n=1}^{\infty}\left(\ln^{n-1}\left(\frac{1}{n+1}\right)-\sin^{-1}\left(\frac{1}{n+2}\right)\right) $$

1. **Finding the $$ n $$th partial sum $$ S_n $$**:
   This series does not telescope in a straightforward manner, and we need to analyze the behavior of the terms as $$ n \to \infty $$.

2. **Behavior of terms**:
   As $$ n \to \infty $$:
   $$
   \ln^{n-1}\left(\frac{1}{n+1}\right) \to -\infty \text{ and } \sin^{-1}\left(\frac{1}{n+2}\right) \to 0
   $$
   Thus, the terms diverge negatively.

3. **Conclusion**:
   The series diverges.

### Summary of Results:
- **39**: Converges, sum = $$ 1 $$
- **40**: Converges, sum = $$ 6 $$
- **41**: Diverges
- **42**: Diverges
- **43**: Diverges

If you need further assistance with any specific series or additional details, feel free to ask!
- **39**: Converges to 1 - **40**: Converges to 6 - **41**: Diverges - **42**: Diverges - **43**: Diverges

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