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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**