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Determine whether the sequence converges or diverges. Exponential sequence and power sequence.

0

$begingroup$

I have this sequence:

$$a_n = frac4^n1 + 9^n$$

How do I figure out if this diverges or converges? I tried using L'Hospital but that seems too complicated here. What else can I do?

sequences-and-series

share|cite|improve this question

edited Mar 27 at 16:35

Kitty Capital

asked Mar 27 at 16:24

Kitty CapitalKitty Capital

1116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You shouldn't piggyback additional questions to your original question, especially since the original question has been answered. Instead, you should ask a second question.
  $endgroup$
  – MPW
  Mar 27 at 16:32
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I have this sequence:

$$a_n = frac4^n1 + 9^n$$

How do I figure out if this diverges or converges? I tried using L'Hospital but that seems too complicated here. What else can I do?

sequences-and-series

share|cite|improve this question

edited Mar 27 at 16:35

Kitty Capital

asked Mar 27 at 16:24

Kitty CapitalKitty Capital

1116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You shouldn't piggyback additional questions to your original question, especially since the original question has been answered. Instead, you should ask a second question.
  $endgroup$
  – MPW
  Mar 27 at 16:32
  
  

  
  
  
  

add a comment | 

$begingroup$

I have this sequence:

$$a_n = frac4^n1 + 9^n$$

How do I figure out if this diverges or converges? I tried using L'Hospital but that seems too complicated here. What else can I do?

sequences-and-series

share|cite|improve this question

edited Mar 27 at 16:35

Kitty Capital

asked Mar 27 at 16:24

Kitty CapitalKitty Capital

1116

$endgroup$

share|cite|improve this question

edited Mar 27 at 16:35

Kitty Capital

asked Mar 27 at 16:24

Kitty CapitalKitty Capital

1116

share|cite|improve this question

edited Mar 27 at 16:35

Kitty Capital

asked Mar 27 at 16:24

Kitty CapitalKitty Capital

1116

asked Mar 27 at 16:24

Kitty CapitalKitty Capital

1116

asked Mar 27 at 16:24

Kitty CapitalKitty Capital

1116

4 Answers
4

active

oldest

votes

1

$begingroup$

You can rewrite this (dividing by $4^n$ on top and bottom) as
$$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$

As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.

Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.

share|cite|improve this answer

answered Mar 27 at 16:31

MPWMPW

31.2k12157

$endgroup$

add a comment | 

1

$begingroup$

It converges to $0$:

$$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$

because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.

share|cite|improve this answer

edited Mar 27 at 16:35

answered Mar 27 at 16:27

st.mathst.math

1,268115

$endgroup$

add a comment | 

0

$begingroup$

It converges and has limit $0$ because $a_n$ is non-negative and

$$
a_n lt frac4^n9^n=left(frac49right)^n
$$
and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.

share|cite|improve this answer

answered Mar 27 at 17:09

PM.PM.

3,5082925

$endgroup$

add a comment | 

0

$begingroup$

For fun:

Let $n ge 1$:

$0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $

$left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.

The limit $n rightarrow infty$ is ?

Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:40

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

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1

$begingroup$

You can rewrite this (dividing by $4^n$ on top and bottom) as
$$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$

As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.

Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.

share|cite|improve this answer

answered Mar 27 at 16:31

MPWMPW

31.2k12157

$endgroup$

add a comment | 

1

$begingroup$

You can rewrite this (dividing by $4^n$ on top and bottom) as
$$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$

As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.

Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.

share|cite|improve this answer

answered Mar 27 at 16:31

MPWMPW

31.2k12157

$endgroup$

add a comment | 

$begingroup$

You can rewrite this (dividing by $4^n$ on top and bottom) as
$$frac1underbrace4^-n_to 0 + underbrace2.25^n_toinfty$$

As $ntoinfty$, the first term in the denominator shrinks to zero and the other term grows without bound; so their sum grows without bound.

Since the denominator grows without bound and the numerator is a nonzero constant, the whole fraction shrinks to zero.

share|cite|improve this answer

answered Mar 27 at 16:31

MPWMPW

31.2k12157

$endgroup$

share|cite|improve this answer

answered Mar 27 at 16:31

MPWMPW

31.2k12157

answered Mar 27 at 16:31

MPWMPW

31.2k12157

answered Mar 27 at 16:31

MPWMPW

31.2k12157

1

$begingroup$

It converges to $0$:

$$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$

because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.

share|cite|improve this answer

edited Mar 27 at 16:35

answered Mar 27 at 16:27

st.mathst.math

1,268115

$endgroup$

add a comment | 

1

$begingroup$

It converges to $0$:

$$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$

because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.

share|cite|improve this answer

edited Mar 27 at 16:35

answered Mar 27 at 16:27

st.mathst.math

1,268115

$endgroup$

add a comment | 

$begingroup$

It converges to $0$:

$$a_n=frac4^-n4^-na_n=frac1(1/4)^n+(9/4)^nto0$$

because $(1/4)^nto 0$ and $(9/4)^nto+infty$ for $ntoinfty$.

share|cite|improve this answer

edited Mar 27 at 16:35

answered Mar 27 at 16:27

st.mathst.math

1,268115

$endgroup$

share|cite|improve this answer

edited Mar 27 at 16:35

answered Mar 27 at 16:27

st.mathst.math

1,268115

answered Mar 27 at 16:27

st.mathst.math

1,268115

answered Mar 27 at 16:27

st.mathst.math

1,268115

0

$begingroup$

It converges and has limit $0$ because $a_n$ is non-negative and

$$
a_n lt frac4^n9^n=left(frac49right)^n
$$
and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.

share|cite|improve this answer

answered Mar 27 at 17:09

PM.PM.

3,5082925

$endgroup$

add a comment | 

0

$begingroup$

It converges and has limit $0$ because $a_n$ is non-negative and

$$
a_n lt frac4^n9^n=left(frac49right)^n
$$
and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.

share|cite|improve this answer

answered Mar 27 at 17:09

PM.PM.

3,5082925

$endgroup$

add a comment | 

$begingroup$

It converges and has limit $0$ because $a_n$ is non-negative and

$$
a_n lt frac4^n9^n=left(frac49right)^n
$$
and the latter is a basic null sequence i.e. $left(frac49right)^n to0$ as $n to infty$.

share|cite|improve this answer

answered Mar 27 at 17:09

PM.PM.

3,5082925

$endgroup$

share|cite|improve this answer

answered Mar 27 at 17:09

PM.PM.

3,5082925

answered Mar 27 at 17:09

PM.PM.

3,5082925

answered Mar 27 at 17:09

PM.PM.

3,5082925

0

$begingroup$

For fun:

Let $n ge 1$:

$0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $

$left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.

The limit $n rightarrow infty$ is ?

Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:40

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

0

$begingroup$

For fun:

Let $n ge 1$:

$0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $

$left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.

The limit $n rightarrow infty$ is ?

Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:40

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

$begingroup$

For fun:

Let $n ge 1$:

$0< dfrac4^n1+9^n lt dfrac4^n9^n =left (dfrac49right )^n lt $

$left (dfrac12 right )^n =dfrac12^n lt dfrac11+n lt dfrac 1n$.

The limit $n rightarrow infty$ is ?

Used: $2^n=(1+1)^n gt 1+ n$ (Binomial expansion).

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:40

Peter SzilasPeter Szilas

12k2822

$endgroup$

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:40

Peter SzilasPeter Szilas

12k2822

answered Mar 27 at 17:40

Peter SzilasPeter Szilas

12k2822

answered Mar 27 at 17:40

Peter SzilasPeter Szilas

12k2822