Divergence of the sequence ${frac{e^n}{2^n+1}}$
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While studying sequences I came across $a_n= frac{e^n}{2^n+1}$ which is divergent according to the answer key.
Attempting to take the limit results in $lim_{n to infty}frac{e^n/2^n}{1+1/2^n}$.
Am I correct in thinking that the reason it is divergent is because $e^n$ grows significantly faster than $2^n$ as $n$ goes to infinity?
Is the same applicable for any ${c^n}/{d^n}$ where $c>d$ and both positive?
calculus sequences-and-series limits
edited Dec 9 '18 at 5:42
Andrews
3901317
asked Dec 9 '18 at 5:13
SolidSnackDriveSolidSnackDrive
1828
$endgroup$
add a comment |
$begingroup$
While studying sequences I came across $a_n= frac{e^n}{2^n+1}$ which is divergent according to the answer key.
Attempting to take the limit results in $lim_{n to infty}frac{e^n/2^n}{1+1/2^n}$.
Am I correct in thinking that the reason it is divergent is because $e^n$ grows significantly faster than $2^n$ as $n$ goes to infinity?
Is the same applicable for any ${c^n}/{d^n}$ where $c>d$ and both positive?
calculus sequences-and-series limits
edited Dec 9 '18 at 5:42
Andrews
3901317
asked Dec 9 '18 at 5:13
SolidSnackDriveSolidSnackDrive
1828
$endgroup$
1
$begingroup$
If $c>d$ then $(c/d)^n = (1+a)^n$ with $a>0$ and $(1+a)^n > 1 + na $ by Bernoulli's inequality
$endgroup$
– RRL
Dec 9 '18 at 5:30
$begingroup$
It diverges because 1 < e/2.
$endgroup$
– William Elliot
Dec 9 '18 at 6:01
add a comment |
$begingroup$
While studying sequences I came across $a_n= frac{e^n}{2^n+1}$ which is divergent according to the answer key.
Attempting to take the limit results in $lim_{n to infty}frac{e^n/2^n}{1+1/2^n}$.
Am I correct in thinking that the reason it is divergent is because $e^n$ grows significantly faster than $2^n$ as $n$ goes to infinity?
Is the same applicable for any ${c^n}/{d^n}$ where $c>d$ and both positive?
calculus sequences-and-series limits
edited Dec 9 '18 at 5:42
Andrews
3901317
asked Dec 9 '18 at 5:13
SolidSnackDriveSolidSnackDrive
1828
$endgroup$
While studying sequences I came across $a_n= frac{e^n}{2^n+1}$ which is divergent according to the answer key.
Attempting to take the limit results in $lim_{n to infty}frac{e^n/2^n}{1+1/2^n}$.
Am I correct in thinking that the reason it is divergent is because $e^n$ grows significantly faster than $2^n$ as $n$ goes to infinity?
Is the same applicable for any ${c^n}/{d^n}$ where $c>d$ and both positive?
calculus sequences-and-series limits
calculus sequences-and-series limits
edited Dec 9 '18 at 5:42
Andrews
3901317
asked Dec 9 '18 at 5:13
SolidSnackDriveSolidSnackDrive
1828
edited Dec 9 '18 at 5:42
Andrews
3901317
asked Dec 9 '18 at 5:13
SolidSnackDriveSolidSnackDrive
1828
edited Dec 9 '18 at 5:42
Andrews
3901317
edited Dec 9 '18 at 5:42
Andrews
3901317
edited Dec 9 '18 at 5:42
Andrews
3901317
3901317
asked Dec 9 '18 at 5:13
SolidSnackDriveSolidSnackDrive
1828
asked Dec 9 '18 at 5:13
SolidSnackDriveSolidSnackDrive
1828
asked Dec 9 '18 at 5:13
SolidSnackDriveSolidSnackDrive
1828
1828
1
$begingroup$
If $c>d$ then $(c/d)^n = (1+a)^n$ with $a>0$ and $(1+a)^n > 1 + na $ by Bernoulli's inequality
$endgroup$
– RRL
Dec 9 '18 at 5:30
$begingroup$
It diverges because 1 < e/2.
$endgroup$
– William Elliot
Dec 9 '18 at 6:01
add a comment |
1
$begingroup$
If $c>d$ then $(c/d)^n = (1+a)^n$ with $a>0$ and $(1+a)^n > 1 + na $ by Bernoulli's inequality
$endgroup$
– RRL
Dec 9 '18 at 5:30
$begingroup$
It diverges because 1 < e/2.
$endgroup$
– William Elliot
Dec 9 '18 at 6:01
1
1
$begingroup$
If $c>d$ then $(c/d)^n = (1+a)^n$ with $a>0$ and $(1+a)^n > 1 + na $ by Bernoulli's inequality
$endgroup$
– RRL
Dec 9 '18 at 5:30
$begingroup$
If $c>d$ then $(c/d)^n = (1+a)^n$ with $a>0$ and $(1+a)^n > 1 + na $ by Bernoulli's inequality
$endgroup$
– RRL
Dec 9 '18 at 5:30
$begingroup$
It diverges because 1 < e/2.
$endgroup$
– William Elliot
Dec 9 '18 at 6:01
$begingroup$
It diverges because 1 < e/2.
$endgroup$
– William Elliot
Dec 9 '18 at 6:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$
From here, a simple comparison is enough:
$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$
answered Dec 9 '18 at 7:00
David PetersonDavid Peterson
8,85621935
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That's a very clean way to prove it, thank you. My original thoughts are also correct though, no?
$endgroup$
– SolidSnackDrive
Dec 9 '18 at 22:47
$begingroup$
Your thoughts are, if c,d are positive which is equivalent to $r>1$
$endgroup$
– David Peterson
Dec 11 '18 at 21:24
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$
From here, a simple comparison is enough:
$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$
answered Dec 9 '18 at 7:00
David PetersonDavid Peterson
8,85621935
$endgroup$
$begingroup$
That's a very clean way to prove it, thank you. My original thoughts are also correct though, no?
$endgroup$
– SolidSnackDrive
Dec 9 '18 at 22:47
$begingroup$
Your thoughts are, if c,d are positive which is equivalent to $r>1$
$endgroup$
– David Peterson
Dec 11 '18 at 21:24
add a comment |
$begingroup$
I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$
From here, a simple comparison is enough:
$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$
answered Dec 9 '18 at 7:00
David PetersonDavid Peterson
8,85621935
$endgroup$
$begingroup$
That's a very clean way to prove it, thank you. My original thoughts are also correct though, no?
$endgroup$
– SolidSnackDrive
Dec 9 '18 at 22:47
$begingroup$
Your thoughts are, if c,d are positive which is equivalent to $r>1$
$endgroup$
– David Peterson
Dec 11 '18 at 21:24
add a comment |
$begingroup$
I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$
From here, a simple comparison is enough:
$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$
answered Dec 9 '18 at 7:00
David PetersonDavid Peterson
8,85621935
$endgroup$
I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$
From here, a simple comparison is enough:
$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$
answered Dec 9 '18 at 7:00
David PetersonDavid Peterson
8,85621935
answered Dec 9 '18 at 7:00
David PetersonDavid Peterson
8,85621935
answered Dec 9 '18 at 7:00
David PetersonDavid Peterson
8,85621935
answered Dec 9 '18 at 7:00
David PetersonDavid Peterson
8,85621935
8,85621935
$begingroup$
That's a very clean way to prove it, thank you. My original thoughts are also correct though, no?
$endgroup$
– SolidSnackDrive
Dec 9 '18 at 22:47
$begingroup$
Your thoughts are, if c,d are positive which is equivalent to $r>1$
$endgroup$
– David Peterson
Dec 11 '18 at 21:24
add a comment |
$begingroup$
That's a very clean way to prove it, thank you. My original thoughts are also correct though, no?
$endgroup$
– SolidSnackDrive
Dec 9 '18 at 22:47
$begingroup$
Your thoughts are, if c,d are positive which is equivalent to $r>1$
$endgroup$
– David Peterson
Dec 11 '18 at 21:24
$begingroup$
That's a very clean way to prove it, thank you. My original thoughts are also correct though, no?
$endgroup$
– SolidSnackDrive
Dec 9 '18 at 22:47
$begingroup$
That's a very clean way to prove it, thank you. My original thoughts are also correct though, no?
$endgroup$
– SolidSnackDrive
Dec 9 '18 at 22:47
$begingroup$
Your thoughts are, if c,d are positive which is equivalent to $r>1$
$endgroup$
– David Peterson
Dec 11 '18 at 21:24
$begingroup$
Your thoughts are, if c,d are positive which is equivalent to $r>1$
$endgroup$
– David Peterson
Dec 11 '18 at 21:24
add a comment |
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1
$begingroup$
If $c>d$ then $(c/d)^n = (1+a)^n$ with $a>0$ and $(1+a)^n > 1 + na $ by Bernoulli's inequality
$endgroup$
– RRL
Dec 9 '18 at 5:30
$begingroup$
It diverges because 1 < e/2.
$endgroup$
– William Elliot
Dec 9 '18 at 6:01