Determine whether the sequence converges or diverges. Exponential sequence and power sequence. The Next CEO of Stack OverflowSeries diverges or convergesDetermine if the sequence converges or diverges.Proving a sequence diverges to infinityDetermine whether or not the series $sumlimits_n=2^infty (fracn+4n+8)^n$ converges or divergesHow to determine whether the series converges or diverges?Proving that a sequence diverges if another one divergesDetermine whether the series converges or diverges.How to determine whether a sequence converges or divergesDetermine whether the sequence converge or divergesDetermine whether the sequence converges or diverges.
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Determine whether the sequence converges or diverges. Exponential sequence and power sequence.
The Next CEO of Stack OverflowSeries diverges or convergesDetermine if the sequence converges or diverges.Proving a sequence diverges to infinityDetermine whether or not the series $sumlimits_n=2^infty (fracn+4n+8)^n$ converges or divergesHow to determine whether the series converges or diverges?Proving that a sequence diverges if another one divergesDetermine whether the series converges or diverges.How to determine whether a sequence converges or divergesDetermine whether the sequence converge or divergesDetermine whether the sequence converges or diverges.
$begingroup$
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
asked 2 days ago
Kitty CapitalKitty Capital
1096
$endgroup$
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
asked 2 days ago
Kitty CapitalKitty Capital
1096
$endgroup$
$begingroup$
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
2 days ago
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
asked 2 days ago
Kitty CapitalKitty Capital
1096
$endgroup$
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
sequences-and-series
asked 2 days ago
Kitty CapitalKitty Capital
1096
asked 2 days ago
Kitty CapitalKitty Capital
1096
edited 2 days ago
Kitty Capital
asked 2 days ago
Kitty CapitalKitty Capital
1096
asked 2 days ago
Kitty CapitalKitty Capital
1096
asked 2 days ago
Kitty CapitalKitty Capital
1096
1096
$begingroup$
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
2 days ago
add a comment |
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.
answered 2 days ago
MPWMPW
31k12157
$endgroup$
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 $ntoinfty$.
answered 2 days ago
st.mathst.math
1,05615
$endgroup$
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 $n to infty$.
answered 2 days ago
PM.PM.
3,4432925
$endgroup$
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).
answered 2 days ago
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 2 days ago
MPWMPW
31k12157
$endgroup$
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.
answered 2 days ago
MPWMPW
31k12157
$endgroup$
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.
answered 2 days ago
MPWMPW
31k12157
$endgroup$
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.
answered 2 days ago
MPWMPW
31k12157
answered 2 days ago
MPWMPW
31k12157
answered 2 days ago
MPWMPW
31k12157
answered 2 days ago
MPWMPW
31k12157
31k12157
add a comment |
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 $ntoinfty$.
answered 2 days ago
st.mathst.math
1,05615
$endgroup$
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 $ntoinfty$.
answered 2 days ago
st.mathst.math
1,05615
$endgroup$
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 $ntoinfty$.
answered 2 days ago
st.mathst.math
1,05615
$endgroup$
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 $ntoinfty$.
answered 2 days ago
st.mathst.math
1,05615
edited 2 days ago
answered 2 days ago
st.mathst.math
1,05615
answered 2 days ago
st.mathst.math
1,05615
answered 2 days ago
st.mathst.math
1,05615
1,05615
add a comment |
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 $n to infty$.
answered 2 days ago
PM.PM.
3,4432925
$endgroup$
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 $n to infty$.
answered 2 days ago
PM.PM.
3,4432925
$endgroup$
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 $n to infty$.
answered 2 days ago
PM.PM.
3,4432925
$endgroup$
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 $n to infty$.
answered 2 days ago
PM.PM.
3,4432925
answered 2 days ago
PM.PM.
3,4432925
answered 2 days ago
PM.PM.
3,4432925
answered 2 days ago
PM.PM.
3,4432925
3,4432925
add a comment |
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).
answered 2 days ago
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
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).
answered 2 days ago
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
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).
answered 2 days ago
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
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).
answered 2 days ago
Peter SzilasPeter Szilas
11.6k2822
edited 2 days ago
answered 2 days ago
Peter SzilasPeter Szilas
11.6k2822
answered 2 days ago
Peter SzilasPeter Szilas
11.6k2822
answered 2 days ago
Peter SzilasPeter Szilas
11.6k2822
11.6k2822
add a comment |
add a comment |
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$begingroup$
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
2 days ago