Mixing
Why does $lim_{nrightarrow infty }left(1+frac{cos(npi )}{n}right)^{n}$ NOT exist?
2
$begingroup$
Why doesn't the following limit exist?
$$lim_{nrightarrow infty }left(1+frac{cos(npi )}{n}right)^{n}$$
I write that because $cos$ has values in $[-1,1]$ and $nrightarrow infty$ this limit can't exists but I don't know if I'm right.
calculus sequences-and-series limits
asked Feb 2 at 15:55
DaniVajaDaniVaja
1027
$endgroup$
add a comment |
2
$begingroup$
Why doesn't the following limit exist?
$$lim_{nrightarrow infty }left(1+frac{cos(npi )}{n}right)^{n}$$
I write that because $cos$ has values in $[-1,1]$ and $nrightarrow infty$ this limit can't exists but I don't know if I'm right.
calculus sequences-and-series limits
asked Feb 2 at 15:55
DaniVajaDaniVaja
1027
$endgroup$
$begingroup$
The expression amounts to $left(1pmfrac1nright)^n$ which tends to $e^{pm1}$.
$endgroup$
– Yves Daoust
Feb 2 at 16:12
add a comment |
2
2
2
$begingroup$
Why doesn't the following limit exist?
$$lim_{nrightarrow infty }left(1+frac{cos(npi )}{n}right)^{n}$$
I write that because $cos$ has values in $[-1,1]$ and $nrightarrow infty$ this limit can't exists but I don't know if I'm right.
calculus sequences-and-series limits
asked Feb 2 at 15:55
DaniVajaDaniVaja
1027
$endgroup$
Why doesn't the following limit exist?
$$lim_{nrightarrow infty }left(1+frac{cos(npi )}{n}right)^{n}$$
I write that because $cos$ has values in $[-1,1]$ and $nrightarrow infty$ this limit can't exists but I don't know if I'm right.
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Feb 2 at 15:55
DaniVajaDaniVaja
1027
asked Feb 2 at 15:55
DaniVajaDaniVaja
1027
edited Feb 2 at 16:09
user587192
asked Feb 2 at 15:55
DaniVajaDaniVaja
1027
asked Feb 2 at 15:55
DaniVajaDaniVaja
1027
asked Feb 2 at 15:55
DaniVajaDaniVaja
1027
1027
$begingroup$
The expression amounts to $left(1pmfrac1nright)^n$ which tends to $e^{pm1}$.
$endgroup$
– Yves Daoust
Feb 2 at 16:12
add a comment |
$begingroup$
The expression amounts to $left(1pmfrac1nright)^n$ which tends to $e^{pm1}$.
$endgroup$
– Yves Daoust
Feb 2 at 16:12
$begingroup$
The expression amounts to $left(1pmfrac1nright)^n$ which tends to $e^{pm1}$.
$endgroup$
– Yves Daoust
Feb 2 at 16:12
$begingroup$
The expression amounts to $left(1pmfrac1nright)^n$ which tends to $e^{pm1}$.
$endgroup$
– Yves Daoust
Feb 2 at 16:12
add a comment |
2 Answers
2
active
oldest
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4
$begingroup$
$cos (n pi)=(-1)^n $. Now look at the subsequences $(a_{2n}) $ and $(a_{2n+1}) $.
answered Feb 2 at 16:00
FredFred
48.5k11849
$endgroup$
$begingroup$
Thank you for your help :)
$endgroup$
– DaniVaja
Feb 2 at 16:39
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct ?
$endgroup$
– DaniVaja
Feb 2 at 16:43
add a comment |
4
$begingroup$
Hint:
A sequence $a_n$ convergence to $L$ if and only if every sub-sequence convergence to $L$.
Consider the sub-sequences $a_{2n},a_{2n+1}$ of $a_n =(1+frac{cos(npi )}{n})^{n}$.
edited Feb 2 at 16:07
Bernard
124k741117
answered Feb 2 at 15:58
YankoYanko
8,4692830
$endgroup$
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct?
$endgroup$
– DaniVaja
Feb 2 at 16:39
1
$begingroup$
@DaniVaja yes that's exactly the idea.
$endgroup$
– Yanko
Feb 3 at 11:02
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
4
$begingroup$
$cos (n pi)=(-1)^n $. Now look at the subsequences $(a_{2n}) $ and $(a_{2n+1}) $.
answered Feb 2 at 16:00
FredFred
48.5k11849
$endgroup$
$begingroup$
Thank you for your help :)
$endgroup$
– DaniVaja
Feb 2 at 16:39
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct ?
$endgroup$
– DaniVaja
Feb 2 at 16:43
add a comment |
4
$begingroup$
$cos (n pi)=(-1)^n $. Now look at the subsequences $(a_{2n}) $ and $(a_{2n+1}) $.
answered Feb 2 at 16:00
FredFred
48.5k11849
$endgroup$
$begingroup$
Thank you for your help :)
$endgroup$
– DaniVaja
Feb 2 at 16:39
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct ?
$endgroup$
– DaniVaja
Feb 2 at 16:43
add a comment |
4
4
4
$begingroup$
$cos (n pi)=(-1)^n $. Now look at the subsequences $(a_{2n}) $ and $(a_{2n+1}) $.
answered Feb 2 at 16:00
FredFred
48.5k11849
$endgroup$
$cos (n pi)=(-1)^n $. Now look at the subsequences $(a_{2n}) $ and $(a_{2n+1}) $.
answered Feb 2 at 16:00
FredFred
48.5k11849
answered Feb 2 at 16:00
FredFred
48.5k11849
answered Feb 2 at 16:00
FredFred
48.5k11849
answered Feb 2 at 16:00
FredFred
48.5k11849
48.5k11849
$begingroup$
Thank you for your help :)
$endgroup$
– DaniVaja
Feb 2 at 16:39
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct ?
$endgroup$
– DaniVaja
Feb 2 at 16:43
add a comment |
$begingroup$
Thank you for your help :)
$endgroup$
– DaniVaja
Feb 2 at 16:39
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct ?
$endgroup$
– DaniVaja
Feb 2 at 16:43
$begingroup$
Thank you for your help :)
$endgroup$
– DaniVaja
Feb 2 at 16:39
$begingroup$
Thank you for your help :)
$endgroup$
– DaniVaja
Feb 2 at 16:39
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct ?
$endgroup$
– DaniVaja
Feb 2 at 16:43
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct ?
$endgroup$
– DaniVaja
Feb 2 at 16:43
add a comment |
4
$begingroup$
Hint:
A sequence $a_n$ convergence to $L$ if and only if every sub-sequence convergence to $L$.
Consider the sub-sequences $a_{2n},a_{2n+1}$ of $a_n =(1+frac{cos(npi )}{n})^{n}$.
edited Feb 2 at 16:07
Bernard
124k741117
answered Feb 2 at 15:58
YankoYanko
8,4692830
$endgroup$
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct?
$endgroup$
– DaniVaja
Feb 2 at 16:39
1
$begingroup$
@DaniVaja yes that's exactly the idea.
$endgroup$
– Yanko
Feb 3 at 11:02
add a comment |
4
$begingroup$
Hint:
A sequence $a_n$ convergence to $L$ if and only if every sub-sequence convergence to $L$.
Consider the sub-sequences $a_{2n},a_{2n+1}$ of $a_n =(1+frac{cos(npi )}{n})^{n}$.
edited Feb 2 at 16:07
Bernard
124k741117
answered Feb 2 at 15:58
YankoYanko
8,4692830
$endgroup$
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct?
$endgroup$
– DaniVaja
Feb 2 at 16:39
1
$begingroup$
@DaniVaja yes that's exactly the idea.
$endgroup$
– Yanko
Feb 3 at 11:02
add a comment |
4
4
4
$begingroup$
Hint:
A sequence $a_n$ convergence to $L$ if and only if every sub-sequence convergence to $L$.
Consider the sub-sequences $a_{2n},a_{2n+1}$ of $a_n =(1+frac{cos(npi )}{n})^{n}$.
edited Feb 2 at 16:07
Bernard
124k741117
answered Feb 2 at 15:58
YankoYanko
8,4692830
$endgroup$
Hint:
A sequence $a_n$ convergence to $L$ if and only if every sub-sequence convergence to $L$.
Consider the sub-sequences $a_{2n},a_{2n+1}$ of $a_n =(1+frac{cos(npi )}{n})^{n}$.
edited Feb 2 at 16:07
Bernard
124k741117
answered Feb 2 at 15:58
YankoYanko
8,4692830
edited Feb 2 at 16:07
Bernard
124k741117
edited Feb 2 at 16:07
Bernard
124k741117
edited Feb 2 at 16:07
Bernard
124k741117
124k741117
answered Feb 2 at 15:58
YankoYanko
8,4692830
answered Feb 2 at 15:58
YankoYanko
8,4692830
answered Feb 2 at 15:58
YankoYanko
8,4692830
8,4692830
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct?
$endgroup$
– DaniVaja
Feb 2 at 16:39
1
$begingroup$
@DaniVaja yes that's exactly the idea.
$endgroup$
– Yanko
Feb 3 at 11:02
add a comment |
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct?
$endgroup$
– DaniVaja
Feb 2 at 16:39
1
$begingroup$
@DaniVaja yes that's exactly the idea.
$endgroup$
– Yanko
Feb 3 at 11:02
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct?
$endgroup$
– DaniVaja
Feb 2 at 16:39
$begingroup$
So limit of $a_{2n}=(1+frac{cos(2npi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+frac{cos(2npi+pi )}{2n+1})^{2n+1}$ because $cos (n pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct?
$endgroup$
– DaniVaja
Feb 2 at 16:39
1
1
$begingroup$
@DaniVaja yes that's exactly the idea.
$endgroup$
– Yanko
Feb 3 at 11:02
$begingroup$
@DaniVaja yes that's exactly the idea.
$endgroup$
– Yanko
Feb 3 at 11:02
add a comment |
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$begingroup$
The expression amounts to $left(1pmfrac1nright)^n$ which tends to $e^{pm1}$.
$endgroup$
– Yves Daoust
Feb 2 at 16:12