Proof of the preliminary test (Divergence test for series)

  • Thread starter joej24
  • Start date
  • #1
78
0

Homework Statement


[tex]Proove \, that \, if \, \lim_{n \to \infty} a_{n} \neq 0 \, , \, \sum^\infty a_{n} \, diverges[/tex]


Homework Equations


[tex] S_{n} - S_{n-1} = a_{n}[/tex]
[tex] \lim_{n \to \infty} S_{n} = S[/tex]

The Attempt at a Solution


[tex] S_{n} - S_{n-1} = a_{n}[/tex]
[tex] \lim_{n \to \infty} S_{n} - S_{n-1} = \lim_{n \to \infty} a_{n}[/tex]
Since [itex] a_{n} \neq 0 [/itex], we can divide by [itex]a_{n}[/itex]

[tex] \lim_{n \to \infty} \frac {S_{n} - S_{n-1}} {a_{n}} = 1[/tex]
We can rewrite [itex] S_{n} [/itex] and [itex] S_{n-1} [/itex] as [itex] \sum^\infty a_{n} [/itex] and [itex] \sum^\infty a_{n-1} [/itex]

So, [tex] \lim_{n \to \infty} \frac {\sum^\infty a_{n} - \sum^\infty a_{n-1}} {a_{n}} = 1[/tex]
Rearranging, [tex] \lim_{n \to \infty} \frac{\sum^\infty a_{n}} {a_{n}} - \sum^\infty \frac{a_{n-1}} {a_{n}} = 1[/tex]
The term inside the second series is 1, so
[tex] \lim_{n \to \infty} \frac{\sum^\infty a_{n}} {a_{n}} = 1 + \sum^\infty 1 [/tex]

Thus, [tex] S_{n} = \lim_{n \to \infty} \sum^\infty a_{n} = a_{n} (1 + \sum^\infty 1) [/tex]
Since [itex] \sum^\infty 1[/itex] diverges, we can say that [itex] S_{n} [/itex] diverges and therefore, if [itex]\lim_{n \to \infty} a_{n} \neq 0 [/itex] , [itex] \sum^\infty a_{n} [/itex] diverges.

Is this proof complete?
 
Last edited:

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,556
767

Homework Statement


[tex]Proove \, that \, if \, \lim_{n \to \infty} a_{n} \neq 0 \, , \, \sum^\infty a_{n} \, diverges[/tex]
That isn't a correct statement of the theorem in the first place. When you say ##\lim_{n \to \infty} a_{n} \neq 0## you are implying ##a_n## has a limit which isn't ##0##. That is not the correct denial of the statement that ##a_n\rightarrow 0##. It may not converge to anything.

Homework Equations


[tex] S_{n} - S_{n-1} = a_{n}[/tex]
[tex] \lim_{n \to \infty} S_{n} = S[/tex]

The Attempt at a Solution


[tex] S_{n} - S_{n-1} = a_{n}[/tex]
[tex] \lim_{n \to \infty} S_{n} - S_{n-1} = \lim_{n \to \infty} a_{n}[/tex]
Since [itex] a_{n} \neq 0 [/itex], we can divide by [itex]a_{n}[/itex]
You can't take a limit since it isn't given ##a_n## has one. And nothing gives you ##a_n\ne 0##.

I would suggest you try the showing the contrapositive: If ##\sum a_n## converges then ##a_n\rightarrow 0##.
 

Related Threads on Proof of the preliminary test (Divergence test for series)

Replies
2
Views
2K
Replies
2
Views
190
Replies
7
Views
1K
  • Last Post
Replies
1
Views
894
  • Last Post
Replies
2
Views
538
Replies
5
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
5
Views
6K
Replies
5
Views
1K
Replies
15
Views
2K
Top