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7.7 The Translation Theorem

7.73   Theorem. Let be a real convergent sequence, say . If for all , then .

Proof: I note that , since if , then . Suppose, to get a contradiction, that , (so ), and let be a precision function for the null sequence . Let . Then , so , and hence . This contradicts the assumption that for all .

7.74   Exercise (Inequality theorem.) A Let be convergent real sequences. Suppose that for all . Prove that .

7.75   Exercise. A Prove the following assertion, or give an example to show that it is not true. Let be convergent real sequences. Suppose that for all . Then .

7.76   Definition (Translate of a sequence.) Let be a sequence and let . Then the sequence is called a translate of .

7.77   Example. If , then . A translate of a sequence is a sequence obtained by ignoring the first few terms.

7.78   Theorem (Translation theorem.) If is a convergent complex sequence, and , then converges, and . Conversely, if converges, then converges to the same limit.

Proof: Let , let and let be a precision function for . I claim is also a precision function for . In fact, for all , and all ,

Conversely, suppose

and let be a precision function for . Let for all . I claim is a precision function for . For all ,

7.79   Example. Let the sequence be defined by

Then

Suppose I knew that converged to a limit . It is clear that for all , so must be . By the translation theorem

so ; i.e., . Hence , and since , we conclude . I've shown that the only thing that can possibly converge to is . Now

Since , we have for all ,

Hence

and by induction,

By theorem 7.64 is a null sequence, and by the comparison theorem for null sequences, it follows that is a null sequence. This completes the proof that .

7.80   Exercise. Let

a)
Assume that converges, and determine the value of .
b)
Calculate , using all of the accuracy of your calculator. Does the sequence appear to converge?

7.81   Entertainment. Show that the sequence defined in the previous exercise converges. We will prove this result in Example 7.97, but you can prove it now, using results you know.

7.82   Exercise. Let be the sequence defined by

a)
Assume that converges, and determine the value of .
b)
Calculate , using all of the accuracy of your calculator. Does this sequence converge?

7.83   Theorem (Divergence test.) Let be complex sequences such that for all . Suppose that and where . Then diverges.

Proof: Suppose, to get a contradiction, that converges to a limit . Then by the product theorem, converges to ; i.e., . This contradicts our assumption that has a non-zero limit.

7.84   Exercise. Prove the following assertion or give an example to show that it is not true: Let be complex sequences such that for all , but . Then diverges.

7.85   Example. Let for all . Then

Since

and

it follows that diverges.

7.86   Exercise. A Let be complex numbers such that for all . Discuss the convergence of . Consider all possible choices for .

Next: 7.8 Bounded Monotonic Sequences Up: 7. Complex Sequences Previous: 7.6 Geometric Series   Index