- #1
roam
- 1,271
- 12
1. Examine the series [tex]\frac{1}{1 . 2} +[/tex][tex]\frac{1}{2 . 3}+[/tex][tex]\frac{1}{3 . 4}+[/tex][tex]\frac{1}{4 . 5}...[/tex] for convergence.
3. The Attempt at a Solution
The following is the book's answer:
"[tex]lim_{n\rightarrow \infty}S_{n}[/tex]
[tex]lim_{n\rightarrow \infty} (1 - \frac{1}{n + 1}) = 1 - 0 = 1[/tex]
Hence the series converge and its sum is 1. "
From [tex]S_{n} = [/tex][tex]\frac{1}{1 . 2} +[/tex][tex]\frac{1}{2 . 3}+[/tex][tex]\frac{1}{3 . 4}+[/tex][tex]\frac{1}{4 . 5}...[/tex]
I can see that the nth term is [tex]\frac{1}{n . (n+1)}[/tex] but I can't follow how the book's obtained the "[tex]1 - \frac{1}{n + 1}[/tex]" or th nth partial sum.
I appreciate some help. Unfortunently there are no explanations in the book on this question.
3. The Attempt at a Solution
The following is the book's answer:
"[tex]lim_{n\rightarrow \infty}S_{n}[/tex]
[tex]lim_{n\rightarrow \infty} (1 - \frac{1}{n + 1}) = 1 - 0 = 1[/tex]
Hence the series converge and its sum is 1. "
From [tex]S_{n} = [/tex][tex]\frac{1}{1 . 2} +[/tex][tex]\frac{1}{2 . 3}+[/tex][tex]\frac{1}{3 . 4}+[/tex][tex]\frac{1}{4 . 5}...[/tex]
I can see that the nth term is [tex]\frac{1}{n . (n+1)}[/tex] but I can't follow how the book's obtained the "[tex]1 - \frac{1}{n + 1}[/tex]" or th nth partial sum.
I appreciate some help. Unfortunently there are no explanations in the book on this question.