Proving limit of $a^nto0$ for $|a|<1$ without use of logarithms

up vote
2
down vote

favorite

Prove that $a^nto0$ as $ntoÃ¢ÂˆÂž$ for $|a|<1$ without use of logarithms by using properties of the sequence $u_n=|a|^n.$

I've noticed that I should use the subsequence $u_2n$, and the fact that $u_2n=u_n^2$. However, I don't know where to go from here. I'm not familiar with these types of proofs so a hint/solution would be greatly appreciated.

Thank you!

asked Aug 2 at 9:35

RJM

133

add a commentÂ |Â

up vote
2
down vote

favorite

Prove that $a^nto0$ as $ntoÃ¢ÂˆÂž$ for $|a|<1$ without use of logarithms by using properties of the sequence $u_n=|a|^n.$

Thank you!

asked Aug 2 at 9:35

RJM

133

add a commentÂ |Â

up vote
2
down vote

favorite

Prove that $a^nto0$ as $ntoÃ¢ÂˆÂž$ for $|a|<1$ without use of logarithms by using properties of the sequence $u_n=|a|^n.$

Thank you!

asked Aug 2 at 9:35

RJM

133

Prove that $a^nto0$ as $ntoÃ¢ÂˆÂž$ for $|a|<1$ without use of logarithms by using properties of the sequence $u_n=|a|^n.$

Thank you!

asked Aug 2 at 9:35

RJM

133

asked Aug 2 at 9:35

RJM

133

asked Aug 2 at 9:35

RJM

133

asked Aug 2 at 9:35

RJM

133

add a commentÂ |Â

7 Answers
7

active

oldest

votes

up vote
2
down vote

accepted

Just for simplicity consider the case $0<a<1.$ Now consider the sequence $u_n=a^n.$ Show that

$u_n$ is decreasing;

$u_n$ is bounded from below.

Thus we have that $u_n$ is convergent. Let's write $L=lim_n u_n.$ Since $u_2n=u_n^2$ we have that

$$lim_n u_2n=L=L^2=lim_n u_n^2.$$ So $L=L^2$ from where $L=0$ or $L=1.$ The case $L= 1$ is not possible since $u_n<1$ and it is decreasing.

edited Aug 2 at 9:56

answered Aug 2 at 9:43

mfl

22.9k11837

Sorry if I'm missing something obvious, but how is it that the limit of u_n is equal to the limit of u_n^2?
â€“Â RJM
Aug 2 at 9:53

I had a typo. I meant $lim_n u_2n=L=L^2=lim_n u_n^2.$ (If $u_n$ is convergent to $L$ then $L=lim_n u_n=lim_n u_2n=lim_n u_2n+7$ because $u_2n, u_2n+7$ are subsequences of a convergent sequence and thus convergent to the same limit.)
â€“Â mfl
Aug 2 at 9:56

add a commentÂ |Â

up vote
2
down vote

If $|a|<1$, then $1/|a| >1$, hence there is $x>0$ such that $frac1a=1+x$. With Bernoulli we get for $ n in mathbb N$:

$frac1a^n=(1+x)^n ge 1+nx > nx$.

Hence

$|a^n| <frac1nx$.

Conclusion: $a^n to 0$.

answered Aug 2 at 9:53

Fred

37k1237

add a commentÂ |Â

up vote
1
down vote

Assume $0le a<1$ and set $b=frac1a >1$ then by Bernoulli inequality

$$b^n=(1-(b-1))^nge 1-n(b-1) to infty$$

and then

$$a^n=frac1b^nto 0$$

More in general for $|a|<1$ we have that

$$-|a|^nle a^nle |a|^n$$

and then by squeeze theorem we conclude that

$$a^n to 0$$

edited Aug 2 at 10:26

answered Aug 2 at 9:53

gimusi

63.8k73480

add a commentÂ |Â

up vote
1
down vote

For fun:

1) $a_n= |a|^n$ is a strictly decreasing sequence for $ 0 <|a| lt 1$.

2) $a_n$ is bounded below by $0$.

3) Hence convergent.

4)Recursively defined by:

$a_n+1 =|a|a_n$, with $a_1=|a|$.

5) $lim_n rightarrow infty a_n+1 = lim_n rightarrow infty(|a|a_n)$.

With $lim_n rightarrow infty a_n+1 =$

$ lim_n rightarrow infty a_n =:L ge 0$,

we have:

6) $L=|a|L$ with $0 < |a| lt 1 $.

Hence?

edited Aug 2 at 11:34

answered Aug 2 at 11:29

Peter Szilas

7,8402617

add a commentÂ |Â

up vote
0
down vote

Note that $$ |a^n+1| = |a||a^n|$$

The sequence $a^n$ is decreasing and bounded below so it converges to $L$

Taking limit of $$ |a^n+1| = |a||a^n|$$ we get $|a|L=L$ which implies $L=0$

answered Aug 2 at 9:57

Mohammad Riazi-Kermani

27.1k41851

add a commentÂ |Â

up vote
0
down vote

Why not just (assuming as mfl did that $0<a<1$) use definitions:
$a^n$ is decreasing and bounded below, hence convergent, as in mfl's post, and the limit $L = inf a^n$ is pretty clearly nonnegative. Assume by way of contradiction that $L>0$. Then there exists $N>0$ such that $nge N$ implies $a^n < L(1/a)$. Then $a^N+1 < L$, contradiction.

answered Aug 2 at 10:01

John Brevik

1,22748

add a commentÂ |Â

up vote
0
down vote

By ratio test

$$frac^n+1=|a|<1 implies |a|^n to 0$$

then since

$$-|a|^nle a^nle |a|^n$$

by squeeze theorem we conclude that

$$a^n to 0$$

answered Aug 2 at 10:13

gimusi

63.8k73480

add a commentÂ |Â

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869873%2fproving-limit-of-an-to0-for-a1-without-use-of-logarithms%23new-answer', 'question_page');

);

Post as a guest

Name

7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

up vote
2
down vote

accepted

Just for simplicity consider the case $0<a<1.$ Now consider the sequence $u_n=a^n.$ Show that

$u_n$ is decreasing;

$u_n$ is bounded from below.

Thus we have that $u_n$ is convergent. Let's write $L=lim_n u_n.$ Since $u_2n=u_n^2$ we have that

$$lim_n u_2n=L=L^2=lim_n u_n^2.$$ So $L=L^2$ from where $L=0$ or $L=1.$ The case $L= 1$ is not possible since $u_n<1$ and it is decreasing.

edited Aug 2 at 9:56

answered Aug 2 at 9:43

mfl

22.9k11837

Sorry if I'm missing something obvious, but how is it that the limit of u_n is equal to the limit of u_n^2?
â€“Â RJM
Aug 2 at 9:53

I had a typo. I meant $lim_n u_2n=L=L^2=lim_n u_n^2.$ (If $u_n$ is convergent to $L$ then $L=lim_n u_n=lim_n u_2n=lim_n u_2n+7$ because $u_2n, u_2n+7$ are subsequences of a convergent sequence and thus convergent to the same limit.)
â€“Â mfl
Aug 2 at 9:56

add a commentÂ |Â

up vote
2
down vote

accepted

Just for simplicity consider the case $0<a<1.$ Now consider the sequence $u_n=a^n.$ Show that

$u_n$ is decreasing;

$u_n$ is bounded from below.

Thus we have that $u_n$ is convergent. Let's write $L=lim_n u_n.$ Since $u_2n=u_n^2$ we have that

$$lim_n u_2n=L=L^2=lim_n u_n^2.$$ So $L=L^2$ from where $L=0$ or $L=1.$ The case $L= 1$ is not possible since $u_n<1$ and it is decreasing.

edited Aug 2 at 9:56

answered Aug 2 at 9:43

mfl

22.9k11837

Sorry if I'm missing something obvious, but how is it that the limit of u_n is equal to the limit of u_n^2?
â€“Â RJM
Aug 2 at 9:53

I had a typo. I meant $lim_n u_2n=L=L^2=lim_n u_n^2.$ (If $u_n$ is convergent to $L$ then $L=lim_n u_n=lim_n u_2n=lim_n u_2n+7$ because $u_2n, u_2n+7$ are subsequences of a convergent sequence and thus convergent to the same limit.)
â€“Â mfl
Aug 2 at 9:56

add a commentÂ |Â

up vote
2
down vote

accepted

Just for simplicity consider the case $0<a<1.$ Now consider the sequence $u_n=a^n.$ Show that

$u_n$ is decreasing;

$u_n$ is bounded from below.

Thus we have that $u_n$ is convergent. Let's write $L=lim_n u_n.$ Since $u_2n=u_n^2$ we have that

$$lim_n u_2n=L=L^2=lim_n u_n^2.$$ So $L=L^2$ from where $L=0$ or $L=1.$ The case $L= 1$ is not possible since $u_n<1$ and it is decreasing.

edited Aug 2 at 9:56

answered Aug 2 at 9:43

mfl

22.9k11837

Just for simplicity consider the case $0<a<1.$ Now consider the sequence $u_n=a^n.$ Show that

$u_n$ is decreasing;

$u_n$ is bounded from below.

Thus we have that $u_n$ is convergent. Let's write $L=lim_n u_n.$ Since $u_2n=u_n^2$ we have that

$$lim_n u_2n=L=L^2=lim_n u_n^2.$$ So $L=L^2$ from where $L=0$ or $L=1.$ The case $L= 1$ is not possible since $u_n<1$ and it is decreasing.

edited Aug 2 at 9:56

answered Aug 2 at 9:43

mfl

22.9k11837

edited Aug 2 at 9:56

answered Aug 2 at 9:43

mfl

22.9k11837

answered Aug 2 at 9:43

mfl

22.9k11837

answered Aug 2 at 9:43

mfl

22.9k11837

Sorry if I'm missing something obvious, but how is it that the limit of u_n is equal to the limit of u_n^2?
â€“Â RJM
Aug 2 at 9:53

I had a typo. I meant $lim_n u_2n=L=L^2=lim_n u_n^2.$ (If $u_n$ is convergent to $L$ then $L=lim_n u_n=lim_n u_2n=lim_n u_2n+7$ because $u_2n, u_2n+7$ are subsequences of a convergent sequence and thus convergent to the same limit.)
â€“Â mfl
Aug 2 at 9:56

add a commentÂ |Â

Sorry if I'm missing something obvious, but how is it that the limit of u_n is equal to the limit of u_n^2?
â€“Â RJM
Aug 2 at 9:53

I had a typo. I meant $lim_n u_2n=L=L^2=lim_n u_n^2.$ (If $u_n$ is convergent to $L$ then $L=lim_n u_n=lim_n u_2n=lim_n u_2n+7$ because $u_2n, u_2n+7$ are subsequences of a convergent sequence and thus convergent to the same limit.)
â€“Â mfl
Aug 2 at 9:56

Sorry if I'm missing something obvious, but how is it that the limit of u_n is equal to the limit of u_n^2?
â€“Â RJM
Aug 2 at 9:53

I had a typo. I meant $lim_n u_2n=L=L^2=lim_n u_n^2.$ (If $u_n$ is convergent to $L$ then $L=lim_n u_n=lim_n u_2n=lim_n u_2n+7$ because $u_2n, u_2n+7$ are subsequences of a convergent sequence and thus convergent to the same limit.)
â€“Â mfl
Aug 2 at 9:56

add a commentÂ |Â

up vote
2
down vote

If $|a|<1$, then $1/|a| >1$, hence there is $x>0$ such that $frac1a=1+x$. With Bernoulli we get for $ n in mathbb N$:

$frac1a^n=(1+x)^n ge 1+nx > nx$.

Hence

$|a^n| <frac1nx$.

Conclusion: $a^n to 0$.

answered Aug 2 at 9:53

Fred

37k1237

add a commentÂ |Â

up vote
2
down vote

If $|a|<1$, then $1/|a| >1$, hence there is $x>0$ such that $frac1a=1+x$. With Bernoulli we get for $ n in mathbb N$:

$frac1a^n=(1+x)^n ge 1+nx > nx$.

Hence

$|a^n| <frac1nx$.

Conclusion: $a^n to 0$.

answered Aug 2 at 9:53

Fred

37k1237

add a commentÂ |Â

up vote
2
down vote

If $|a|<1$, then $1/|a| >1$, hence there is $x>0$ such that $frac1a=1+x$. With Bernoulli we get for $ n in mathbb N$:

$frac1a^n=(1+x)^n ge 1+nx > nx$.

Hence

$|a^n| <frac1nx$.

Conclusion: $a^n to 0$.

answered Aug 2 at 9:53

Fred

37k1237

If $|a|<1$, then $1/|a| >1$, hence there is $x>0$ such that $frac1a=1+x$. With Bernoulli we get for $ n in mathbb N$:

$frac1a^n=(1+x)^n ge 1+nx > nx$.

Hence

$|a^n| <frac1nx$.

Conclusion: $a^n to 0$.

answered Aug 2 at 9:53

Fred

37k1237

answered Aug 2 at 9:53

Fred

37k1237

answered Aug 2 at 9:53

Fred

37k1237

answered Aug 2 at 9:53

Fred

37k1237

add a commentÂ |Â

up vote
1
down vote

Assume $0le a<1$ and set $b=frac1a >1$ then by Bernoulli inequality

$$b^n=(1-(b-1))^nge 1-n(b-1) to infty$$

and then

$$a^n=frac1b^nto 0$$

More in general for $|a|<1$ we have that

$$-|a|^nle a^nle |a|^n$$

and then by squeeze theorem we conclude that

$$a^n to 0$$

edited Aug 2 at 10:26

answered Aug 2 at 9:53

gimusi

63.8k73480

add a commentÂ |Â

up vote
1
down vote

Assume $0le a<1$ and set $b=frac1a >1$ then by Bernoulli inequality

$$b^n=(1-(b-1))^nge 1-n(b-1) to infty$$

and then

$$a^n=frac1b^nto 0$$

More in general for $|a|<1$ we have that

$$-|a|^nle a^nle |a|^n$$

and then by squeeze theorem we conclude that

$$a^n to 0$$

edited Aug 2 at 10:26

answered Aug 2 at 9:53

gimusi

63.8k73480

add a commentÂ |Â

up vote
1
down vote

Assume $0le a<1$ and set $b=frac1a >1$ then by Bernoulli inequality

$$b^n=(1-(b-1))^nge 1-n(b-1) to infty$$

and then

$$a^n=frac1b^nto 0$$

More in general for $|a|<1$ we have that

$$-|a|^nle a^nle |a|^n$$

and then by squeeze theorem we conclude that

$$a^n to 0$$

edited Aug 2 at 10:26

answered Aug 2 at 9:53

gimusi

63.8k73480

Assume $0le a<1$ and set $b=frac1a >1$ then by Bernoulli inequality

$$b^n=(1-(b-1))^nge 1-n(b-1) to infty$$

and then

$$a^n=frac1b^nto 0$$

More in general for $|a|<1$ we have that

$$-|a|^nle a^nle |a|^n$$

and then by squeeze theorem we conclude that

$$a^n to 0$$

edited Aug 2 at 10:26

answered Aug 2 at 9:53

gimusi

63.8k73480

edited Aug 2 at 10:26

answered Aug 2 at 9:53

gimusi

63.8k73480

answered Aug 2 at 9:53

gimusi

63.8k73480

answered Aug 2 at 9:53

gimusi

63.8k73480

add a commentÂ |Â

up vote
1
down vote

For fun:

1) $a_n= |a|^n$ is a strictly decreasing sequence for $ 0 <|a| lt 1$.

2) $a_n$ is bounded below by $0$.

3) Hence convergent.

4)Recursively defined by:

$a_n+1 =|a|a_n$, with $a_1=|a|$.

5) $lim_n rightarrow infty a_n+1 = lim_n rightarrow infty(|a|a_n)$.

With $lim_n rightarrow infty a_n+1 =$

$ lim_n rightarrow infty a_n =:L ge 0$,

we have:

6) $L=|a|L$ with $0 < |a| lt 1 $.

Hence?

edited Aug 2 at 11:34

answered Aug 2 at 11:29

Peter Szilas

7,8402617

add a commentÂ |Â

up vote
1
down vote

For fun:

1) $a_n= |a|^n$ is a strictly decreasing sequence for $ 0 <|a| lt 1$.

2) $a_n$ is bounded below by $0$.

3) Hence convergent.

4)Recursively defined by:

$a_n+1 =|a|a_n$, with $a_1=|a|$.

5) $lim_n rightarrow infty a_n+1 = lim_n rightarrow infty(|a|a_n)$.

With $lim_n rightarrow infty a_n+1 =$

$ lim_n rightarrow infty a_n =:L ge 0$,

we have:

6) $L=|a|L$ with $0 < |a| lt 1 $.

Hence?

edited Aug 2 at 11:34

answered Aug 2 at 11:29

Peter Szilas

7,8402617

add a commentÂ |Â

up vote
1
down vote

For fun:

1) $a_n= |a|^n$ is a strictly decreasing sequence for $ 0 <|a| lt 1$.

2) $a_n$ is bounded below by $0$.

3) Hence convergent.

4)Recursively defined by:

$a_n+1 =|a|a_n$, with $a_1=|a|$.

5) $lim_n rightarrow infty a_n+1 = lim_n rightarrow infty(|a|a_n)$.

With $lim_n rightarrow infty a_n+1 =$

$ lim_n rightarrow infty a_n =:L ge 0$,

we have:

6) $L=|a|L$ with $0 < |a| lt 1 $.

Hence?

edited Aug 2 at 11:34

answered Aug 2 at 11:29

Peter Szilas

7,8402617

For fun:

1) $a_n= |a|^n$ is a strictly decreasing sequence for $ 0 <|a| lt 1$.

2) $a_n$ is bounded below by $0$.

3) Hence convergent.

4)Recursively defined by:

$a_n+1 =|a|a_n$, with $a_1=|a|$.

5) $lim_n rightarrow infty a_n+1 = lim_n rightarrow infty(|a|a_n)$.

With $lim_n rightarrow infty a_n+1 =$

$ lim_n rightarrow infty a_n =:L ge 0$,

we have:

6) $L=|a|L$ with $0 < |a| lt 1 $.

Hence?

edited Aug 2 at 11:34

answered Aug 2 at 11:29

Peter Szilas

7,8402617

edited Aug 2 at 11:34

answered Aug 2 at 11:29

Peter Szilas

7,8402617

answered Aug 2 at 11:29

Peter Szilas

7,8402617

answered Aug 2 at 11:29

Peter Szilas

7,8402617

add a commentÂ |Â

up vote
0
down vote

Note that $$ |a^n+1| = |a||a^n|$$

The sequence $a^n$ is decreasing and bounded below so it converges to $L$

Taking limit of $$ |a^n+1| = |a||a^n|$$ we get $|a|L=L$ which implies $L=0$

answered Aug 2 at 9:57

Mohammad Riazi-Kermani

27.1k41851

add a commentÂ |Â

up vote
0
down vote

Note that $$ |a^n+1| = |a||a^n|$$

The sequence $a^n$ is decreasing and bounded below so it converges to $L$

Taking limit of $$ |a^n+1| = |a||a^n|$$ we get $|a|L=L$ which implies $L=0$

answered Aug 2 at 9:57

Mohammad Riazi-Kermani

27.1k41851

add a commentÂ |Â

up vote
0
down vote

Note that $$ |a^n+1| = |a||a^n|$$

The sequence $a^n$ is decreasing and bounded below so it converges to $L$

Taking limit of $$ |a^n+1| = |a||a^n|$$ we get $|a|L=L$ which implies $L=0$

answered Aug 2 at 9:57

Mohammad Riazi-Kermani

27.1k41851

Note that $$ |a^n+1| = |a||a^n|$$

The sequence $a^n$ is decreasing and bounded below so it converges to $L$

Taking limit of $$ |a^n+1| = |a||a^n|$$ we get $|a|L=L$ which implies $L=0$

answered Aug 2 at 9:57

Mohammad Riazi-Kermani

27.1k41851

answered Aug 2 at 9:57

Mohammad Riazi-Kermani

27.1k41851

answered Aug 2 at 9:57

Mohammad Riazi-Kermani

27.1k41851

answered Aug 2 at 9:57

Mohammad Riazi-Kermani

27.1k41851

add a commentÂ |Â

up vote
0
down vote

answered Aug 2 at 10:01

John Brevik

1,22748

add a commentÂ |Â

up vote
0
down vote

answered Aug 2 at 10:01

John Brevik

1,22748

add a commentÂ |Â

up vote
0
down vote

answered Aug 2 at 10:01

John Brevik

1,22748

answered Aug 2 at 10:01

John Brevik

1,22748

answered Aug 2 at 10:01

John Brevik

1,22748

answered Aug 2 at 10:01

John Brevik

1,22748

answered Aug 2 at 10:01

John Brevik

1,22748

add a commentÂ |Â

up vote
0
down vote

By ratio test

$$frac^n+1=|a|<1 implies |a|^n to 0$$

then since

$$-|a|^nle a^nle |a|^n$$

by squeeze theorem we conclude that

$$a^n to 0$$

answered Aug 2 at 10:13

gimusi

63.8k73480

add a commentÂ |Â

up vote
0
down vote

By ratio test

$$frac^n+1=|a|<1 implies |a|^n to 0$$

then since

$$-|a|^nle a^nle |a|^n$$

by squeeze theorem we conclude that

$$a^n to 0$$

answered Aug 2 at 10:13

gimusi

63.8k73480

add a commentÂ |Â

up vote
0
down vote

By ratio test

$$frac^n+1=|a|<1 implies |a|^n to 0$$

then since

$$-|a|^nle a^nle |a|^n$$

by squeeze theorem we conclude that

$$a^n to 0$$

answered Aug 2 at 10:13

gimusi

63.8k73480

By ratio test

$$frac^n+1=|a|<1 implies |a|^n to 0$$

then since

$$-|a|^nle a^nle |a|^n$$

by squeeze theorem we conclude that

$$a^n to 0$$

answered Aug 2 at 10:13

gimusi

63.8k73480

answered Aug 2 at 10:13

gimusi

63.8k73480

answered Aug 2 at 10:13

gimusi

63.8k73480

answered Aug 2 at 10:13

gimusi

63.8k73480

add a commentÂ |Â

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

Search This Blog

ukmuiik