Divergence of the sequence ${frac{e^n}{2^n+1}}$












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$begingroup$


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







share|cite|improve this question











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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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 5:42









Andrews

3901317




3901317










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














  • 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$



From here, a simple comparison is enough:



$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$



From here, a simple comparison is enough:



$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$



From here, a simple comparison is enough:



$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$



From here, a simple comparison is enough:



$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$






share|cite|improve this answer









$endgroup$



I assume you are familiar with the following fact: If $r>1$
$$lim_{ntoinfty} r^n = infty$$



From here, a simple comparison is enough:



$$dfrac{e^n}{2^n+1} > dfrac{e^n}{2^n+2^n} = dfrac{1}{2}cdotleft(dfrac{e}{2}right)^n$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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