Prove that $sqrt[k] frac{1}{k+2} to 1$ by squeeze theorem

1

$begingroup$

I wish to show that for $k to infty$:
$$sqrt[k] frac{1}{k+2} to 1$$ by the squeeze theorem.
I was thinking about the following upper bound:
$$sqrt[k] frac{1}{k+2} leq sqrt[k] frac{k}{k+2}=sqrt[k] frac{1}{1+ frac{2}{k}} leq sqrt[k] frac{1}{1} to 1 $$

I am having trouble with finding a lower bound, I think $sqrt[k] frac{1}{k+k}$ is too loose of a bound, I maybe need something stricter.

real-analysis calculus sequences-and-series limits

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edited Jan 2 at 17:21

Wesley Strik

asked Jan 2 at 17:08

Wesley StrikWesley Strik

1,646423

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think $k+k$ in denominator is too loose of a bound?
  $endgroup$
  – Wojowu
  Jan 2 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not see how $sqrt[k]frac{1}{2k} to 1$
  $endgroup$
  – Wesley Strik
  Jan 2 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer below proves that
  $endgroup$
  – Wojowu
  Jan 2 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  True. I understand it as well now. In the limit we get $frac{1}{2^0}=1$ and $sqrt[k]{k}=1$, very neat.
  $endgroup$
  – Wesley Strik
  Jan 2 at 17:24
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I wish to show that for $k to infty$:
$$sqrt[k] frac{1}{k+2} to 1$$ by the squeeze theorem.
I was thinking about the following upper bound:
$$sqrt[k] frac{1}{k+2} leq sqrt[k] frac{k}{k+2}=sqrt[k] frac{1}{1+ frac{2}{k}} leq sqrt[k] frac{1}{1} to 1 $$

I am having trouble with finding a lower bound, I think $sqrt[k] frac{1}{k+k}$ is too loose of a bound, I maybe need something stricter.

real-analysis calculus sequences-and-series limits

share|cite|improve this question

edited Jan 2 at 17:21

Wesley Strik

asked Jan 2 at 17:08

Wesley StrikWesley Strik

1,646423

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think $k+k$ in denominator is too loose of a bound?
  $endgroup$
  – Wojowu
  Jan 2 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not see how $sqrt[k]frac{1}{2k} to 1$
  $endgroup$
  – Wesley Strik
  Jan 2 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer below proves that
  $endgroup$
  – Wojowu
  Jan 2 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  True. I understand it as well now. In the limit we get $frac{1}{2^0}=1$ and $sqrt[k]{k}=1$, very neat.
  $endgroup$
  – Wesley Strik
  Jan 2 at 17:24
  
  
  

  
  
  
  

add a comment | 

1

1

1

1

$begingroup$

I wish to show that for $k to infty$:
$$sqrt[k] frac{1}{k+2} to 1$$ by the squeeze theorem.
I was thinking about the following upper bound:
$$sqrt[k] frac{1}{k+2} leq sqrt[k] frac{k}{k+2}=sqrt[k] frac{1}{1+ frac{2}{k}} leq sqrt[k] frac{1}{1} to 1 $$

I am having trouble with finding a lower bound, I think $sqrt[k] frac{1}{k+k}$ is too loose of a bound, I maybe need something stricter.

real-analysis calculus sequences-and-series limits

share|cite|improve this question

edited Jan 2 at 17:21

Wesley Strik

asked Jan 2 at 17:08

Wesley StrikWesley Strik

1,646423

$endgroup$

I wish to show that for $k to infty$:
$$sqrt[k] frac{1}{k+2} to 1$$ by the squeeze theorem.
I was thinking about the following upper bound:
$$sqrt[k] frac{1}{k+2} leq sqrt[k] frac{k}{k+2}=sqrt[k] frac{1}{1+ frac{2}{k}} leq sqrt[k] frac{1}{1} to 1 $$

I am having trouble with finding a lower bound, I think $sqrt[k] frac{1}{k+k}$ is too loose of a bound, I maybe need something stricter.

real-analysis calculus sequences-and-series limits

real-analysis calculus sequences-and-series limits

share|cite|improve this question

edited Jan 2 at 17:21

Wesley Strik

asked Jan 2 at 17:08

Wesley StrikWesley Strik

1,646423

share|cite|improve this question

edited Jan 2 at 17:21

Wesley Strik

asked Jan 2 at 17:08

Wesley StrikWesley Strik

1,646423

share|cite|improve this question

share|cite|improve this question

edited Jan 2 at 17:21

Wesley Strik

edited Jan 2 at 17:21

Wesley Strik

edited Jan 2 at 17:21

Wesley Strik

asked Jan 2 at 17:08

Wesley StrikWesley Strik

1,646423

asked Jan 2 at 17:08

Wesley StrikWesley Strik

1,646423

asked Jan 2 at 17:08

Wesley StrikWesley Strik

1,646423

1,646423

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think $k+k$ in denominator is too loose of a bound?
  $endgroup$
  – Wojowu
  Jan 2 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not see how $sqrt[k]frac{1}{2k} to 1$
  $endgroup$
  – Wesley Strik
  Jan 2 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer below proves that
  $endgroup$
  – Wojowu
  Jan 2 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  True. I understand it as well now. In the limit we get $frac{1}{2^0}=1$ and $sqrt[k]{k}=1$, very neat.
  $endgroup$
  – Wesley Strik
  Jan 2 at 17:24
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think $k+k$ in denominator is too loose of a bound?
  $endgroup$
  – Wojowu
  Jan 2 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not see how $sqrt[k]frac{1}{2k} to 1$
  $endgroup$
  – Wesley Strik
  Jan 2 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer below proves that
  $endgroup$
  – Wojowu
  Jan 2 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  True. I understand it as well now. In the limit we get $frac{1}{2^0}=1$ and $sqrt[k]{k}=1$, very neat.
  $endgroup$
  – Wesley Strik
  Jan 2 at 17:24
  
  
  

  
  
  
  

1

1

$begingroup$
Why do you think $k+k$ in denominator is too loose of a bound?
$endgroup$
– Wojowu
Jan 2 at 17:20

$begingroup$
Why do you think $k+k$ in denominator is too loose of a bound?
$endgroup$
– Wojowu
Jan 2 at 17:20

$begingroup$
I do not see how $sqrt[k]frac{1}{2k} to 1$
$endgroup$
– Wesley Strik
Jan 2 at 17:22

$begingroup$
I do not see how $sqrt[k]frac{1}{2k} to 1$
$endgroup$
– Wesley Strik
Jan 2 at 17:22

$begingroup$
The answer below proves that
$endgroup$
– Wojowu
Jan 2 at 17:22

$begingroup$
The answer below proves that
$endgroup$
– Wojowu
Jan 2 at 17:22

$begingroup$
True. I understand it as well now. In the limit we get $frac{1}{2^0}=1$ and $sqrt[k]{k}=1$, very neat.
$endgroup$
– Wesley Strik
Jan 2 at 17:24

$begingroup$
True. I understand it as well now. In the limit we get $frac{1}{2^0}=1$ and $sqrt[k]{k}=1$, very neat.
$endgroup$
– Wesley Strik
Jan 2 at 17:24

add a comment | 

2 Answers
2

active

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3

$begingroup$

Note that for $kgeq 2$
$$
left(frac{1}{k+2}right)^{1/k}geq left(frac{1}{2k}right)^{1/k}=frac{1}{2^{1/k}}timesfrac{1}{k^{1/k}}to1times1=1
$$
as $kto infty$.

share|cite|improve this answer

answered Jan 2 at 17:19

Foobaz JohnFoobaz John

21.6k41352

$endgroup$

add a comment | 

0

$begingroup$

Another squeezing argument follows from
$$
liminf frac{a_{k+1}}{a_k} leqliminf sqrt[k]{a_k} leq limsupsqrt[k]{a_k} leq limsup frac{a_{k+1}}{a_k}.
$$
Letting $displaystyle a_k=frac{1}{k+2}$ implies
$$
liminf frac{k+2}{k+3} leqliminf sqrt[k]{frac{1}{k+2}} leq limsupsqrt[k]{frac{1}{k+2}} leq limsup frac{k+2}{k+3}.
$$
Consequently, $displaystyle limsqrt[k]{frac{1}{k+2}} to 1$.

share|cite|improve this answer

answered Jan 2 at 17:34

BerkheimerBerkheimer

1,437924

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add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

3

$begingroup$

Note that for $kgeq 2$
$$
left(frac{1}{k+2}right)^{1/k}geq left(frac{1}{2k}right)^{1/k}=frac{1}{2^{1/k}}timesfrac{1}{k^{1/k}}to1times1=1
$$
as $kto infty$.

share|cite|improve this answer

answered Jan 2 at 17:19

Foobaz JohnFoobaz John

21.6k41352

$endgroup$

add a comment | 

3

$begingroup$

Note that for $kgeq 2$
$$
left(frac{1}{k+2}right)^{1/k}geq left(frac{1}{2k}right)^{1/k}=frac{1}{2^{1/k}}timesfrac{1}{k^{1/k}}to1times1=1
$$
as $kto infty$.

share|cite|improve this answer

answered Jan 2 at 17:19

Foobaz JohnFoobaz John

21.6k41352

$endgroup$

add a comment | 

3

3

3

$begingroup$

Note that for $kgeq 2$
$$
left(frac{1}{k+2}right)^{1/k}geq left(frac{1}{2k}right)^{1/k}=frac{1}{2^{1/k}}timesfrac{1}{k^{1/k}}to1times1=1
$$
as $kto infty$.

share|cite|improve this answer

answered Jan 2 at 17:19

Foobaz JohnFoobaz John

21.6k41352

$endgroup$

Note that for $kgeq 2$
$$
left(frac{1}{k+2}right)^{1/k}geq left(frac{1}{2k}right)^{1/k}=frac{1}{2^{1/k}}timesfrac{1}{k^{1/k}}to1times1=1
$$
as $kto infty$.

share|cite|improve this answer

answered Jan 2 at 17:19

Foobaz JohnFoobaz John

21.6k41352

share|cite|improve this answer

share|cite|improve this answer

answered Jan 2 at 17:19

Foobaz JohnFoobaz John

21.6k41352

answered Jan 2 at 17:19

Foobaz JohnFoobaz John

21.6k41352

answered Jan 2 at 17:19

Foobaz JohnFoobaz John

21.6k41352

21.6k41352

add a comment | 

add a comment | 

0

$begingroup$

Another squeezing argument follows from
$$
liminf frac{a_{k+1}}{a_k} leqliminf sqrt[k]{a_k} leq limsupsqrt[k]{a_k} leq limsup frac{a_{k+1}}{a_k}.
$$
Letting $displaystyle a_k=frac{1}{k+2}$ implies
$$
liminf frac{k+2}{k+3} leqliminf sqrt[k]{frac{1}{k+2}} leq limsupsqrt[k]{frac{1}{k+2}} leq limsup frac{k+2}{k+3}.
$$
Consequently, $displaystyle limsqrt[k]{frac{1}{k+2}} to 1$.

share|cite|improve this answer

answered Jan 2 at 17:34

BerkheimerBerkheimer

1,437924

$endgroup$

add a comment | 

0

$begingroup$

Another squeezing argument follows from
$$
liminf frac{a_{k+1}}{a_k} leqliminf sqrt[k]{a_k} leq limsupsqrt[k]{a_k} leq limsup frac{a_{k+1}}{a_k}.
$$
Letting $displaystyle a_k=frac{1}{k+2}$ implies
$$
liminf frac{k+2}{k+3} leqliminf sqrt[k]{frac{1}{k+2}} leq limsupsqrt[k]{frac{1}{k+2}} leq limsup frac{k+2}{k+3}.
$$
Consequently, $displaystyle limsqrt[k]{frac{1}{k+2}} to 1$.

share|cite|improve this answer

answered Jan 2 at 17:34

BerkheimerBerkheimer

1,437924

$endgroup$

add a comment | 

0

0

0

$begingroup$

Another squeezing argument follows from
$$
liminf frac{a_{k+1}}{a_k} leqliminf sqrt[k]{a_k} leq limsupsqrt[k]{a_k} leq limsup frac{a_{k+1}}{a_k}.
$$
Letting $displaystyle a_k=frac{1}{k+2}$ implies
$$
liminf frac{k+2}{k+3} leqliminf sqrt[k]{frac{1}{k+2}} leq limsupsqrt[k]{frac{1}{k+2}} leq limsup frac{k+2}{k+3}.
$$
Consequently, $displaystyle limsqrt[k]{frac{1}{k+2}} to 1$.

share|cite|improve this answer

answered Jan 2 at 17:34

BerkheimerBerkheimer

1,437924

$endgroup$

Another squeezing argument follows from
$$
liminf frac{a_{k+1}}{a_k} leqliminf sqrt[k]{a_k} leq limsupsqrt[k]{a_k} leq limsup frac{a_{k+1}}{a_k}.
$$
Letting $displaystyle a_k=frac{1}{k+2}$ implies
$$
liminf frac{k+2}{k+3} leqliminf sqrt[k]{frac{1}{k+2}} leq limsupsqrt[k]{frac{1}{k+2}} leq limsup frac{k+2}{k+3}.
$$
Consequently, $displaystyle limsqrt[k]{frac{1}{k+2}} to 1$.

share|cite|improve this answer

answered Jan 2 at 17:34

BerkheimerBerkheimer

1,437924

share|cite|improve this answer

share|cite|improve this answer

answered Jan 2 at 17:34

BerkheimerBerkheimer

1,437924

answered Jan 2 at 17:34

BerkheimerBerkheimer

1,437924

answered Jan 2 at 17:34

BerkheimerBerkheimer

1,437924

1,437924

add a comment | 

add a comment | 

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