Proof of the preliminary test (Divergence test for series)

In summary, the theorem states that if the series converges, then the terms of the series must approach 0. This can be proven by showing the contrapositive: if the terms of the series do not approach 0, then the series does not converge. This can be shown by taking the limit of the series and using properties of limits to show that it does not converge. Therefore, if the limit of the terms of the series is not 0, the series must diverge.
  • #1
joej24
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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?
 
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  • #2
joej24 said:

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##.
 

FAQ: Proof of the preliminary test (Divergence test for series)

What is the purpose of the divergence test for series?

The divergence test is used to determine whether an infinite series is convergent or divergent. It is often the first test used to determine the convergence or divergence of a series.

How does the divergence test work?

The test checks for the divergence of a series by examining the behavior of its terms. If the limit of the terms of the series does not approach zero, then the series is divergent. If the limit does approach zero, then the test is inconclusive and another test must be used to determine the convergence or divergence of the series.

What is the formula for the divergence test?

The formula for the divergence test is stated as follows: If the limit of the terms of a series is not equal to zero, then the series diverges.

Are there any exceptions to the divergence test?

Yes, there are some exceptions to the divergence test. For example, the test cannot be used on alternating series or series with terms that do not approach zero. In these cases, a different test must be used to determine convergence or divergence.

Can the divergence test be used to determine the exact value of a convergent series?

No, the divergence test only determines whether a series is convergent or divergent. It does not provide the exact value of a convergent series. To find the exact value, other methods such as the integral test or comparison test may be used.

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