Mixing
Limit of $a_n=(1-frac13)^2cdot(1-frac16)^2ldots(1-frac{1}{frac{(n)(n+1)}{2}})^2 ; ;forall n geq 2$
$begingroup$
$$a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2ldotsleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
; ;forall n geq 2$$
I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root
$$lim_{nrightarrowinfty} a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
= lim_{nrightarrowinfty}left[left(1-frac13right)^{2n}cdotleft(1-frac16right)^{2n}cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}right]^{frac1n} = lim_{nrightarrowinfty}left(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}=1 $$
sequences-and-series limits limits-without-lhopital cauchy-sequences
edited Jan 17 at 21:33
user1337
46110
asked Jan 17 at 21:12
AbhayAbhay
3569
$endgroup$
add a comment |
$begingroup$
$$a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2ldotsleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
; ;forall n geq 2$$
I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root
$$lim_{nrightarrowinfty} a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
= lim_{nrightarrowinfty}left[left(1-frac13right)^{2n}cdotleft(1-frac16right)^{2n}cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}right]^{frac1n} = lim_{nrightarrowinfty}left(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}=1 $$
sequences-and-series limits limits-without-lhopital cauchy-sequences
edited Jan 17 at 21:33
user1337
46110
asked Jan 17 at 21:12
AbhayAbhay
3569
$endgroup$
add a comment |
$begingroup$
$$a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2ldotsleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
; ;forall n geq 2$$
I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root
$$lim_{nrightarrowinfty} a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
= lim_{nrightarrowinfty}left[left(1-frac13right)^{2n}cdotleft(1-frac16right)^{2n}cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}right]^{frac1n} = lim_{nrightarrowinfty}left(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}=1 $$
sequences-and-series limits limits-without-lhopital cauchy-sequences
edited Jan 17 at 21:33
user1337
46110
asked Jan 17 at 21:12
AbhayAbhay
3569
$endgroup$
$$a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2ldotsleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
; ;forall n geq 2$$
I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root
$$lim_{nrightarrowinfty} a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
= lim_{nrightarrowinfty}left[left(1-frac13right)^{2n}cdotleft(1-frac16right)^{2n}cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}right]^{frac1n} = lim_{nrightarrowinfty}left(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}=1 $$
sequences-and-series limits limits-without-lhopital cauchy-sequences
sequences-and-series limits limits-without-lhopital cauchy-sequences
edited Jan 17 at 21:33
user1337
46110
asked Jan 17 at 21:12
AbhayAbhay
3569
edited Jan 17 at 21:33
user1337
46110
asked Jan 17 at 21:12
AbhayAbhay
3569
edited Jan 17 at 21:33
user1337
46110
edited Jan 17 at 21:33
user1337
46110
edited Jan 17 at 21:33
user1337
46110
46110
asked Jan 17 at 21:12
AbhayAbhay
3569
asked Jan 17 at 21:12
AbhayAbhay
3569
asked Jan 17 at 21:12
AbhayAbhay
3569
3569
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: $$
1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
$$
answered Jan 17 at 21:18
SongSong
15.1k1636
$endgroup$
$begingroup$
True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
$endgroup$
– Abhay
Jan 17 at 21:20
$begingroup$
Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
$endgroup$
– Nick Peterson
Jan 17 at 21:23
$begingroup$
@learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
$endgroup$
– Henry
Jan 17 at 21:24
add a comment |
$begingroup$
One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.
answered Jan 17 at 21:33
Peter ForemanPeter Foreman
2,399214
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
$begingroup$
Hint: $$
1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
$$
answered Jan 17 at 21:18
SongSong
15.1k1636
$endgroup$
$begingroup$
True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
$endgroup$
– Abhay
Jan 17 at 21:20
$begingroup$
Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
$endgroup$
– Nick Peterson
Jan 17 at 21:23
$begingroup$
@learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
$endgroup$
– Henry
Jan 17 at 21:24
add a comment |
$begingroup$
Hint: $$
1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
$$
answered Jan 17 at 21:18
SongSong
15.1k1636
$endgroup$
$begingroup$
True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
$endgroup$
– Abhay
Jan 17 at 21:20
$begingroup$
Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
$endgroup$
– Nick Peterson
Jan 17 at 21:23
$begingroup$
@learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
$endgroup$
– Henry
Jan 17 at 21:24
add a comment |
$begingroup$
Hint: $$
1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
$$
answered Jan 17 at 21:18
SongSong
15.1k1636
$endgroup$
Hint: $$
1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
$$
answered Jan 17 at 21:18
SongSong
15.1k1636
answered Jan 17 at 21:18
SongSong
15.1k1636
answered Jan 17 at 21:18
SongSong
15.1k1636
answered Jan 17 at 21:18
SongSong
15.1k1636
15.1k1636
$begingroup$
True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
$endgroup$
– Abhay
Jan 17 at 21:20
$begingroup$
Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
$endgroup$
– Nick Peterson
Jan 17 at 21:23
$begingroup$
@learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
$endgroup$
– Henry
Jan 17 at 21:24
add a comment |
$begingroup$
True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
$endgroup$
– Abhay
Jan 17 at 21:20
$begingroup$
Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
$endgroup$
– Nick Peterson
Jan 17 at 21:23
$begingroup$
@learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
$endgroup$
– Henry
Jan 17 at 21:24
$begingroup$
True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
$endgroup$
– Abhay
Jan 17 at 21:20
$begingroup$
True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
$endgroup$
– Abhay
Jan 17 at 21:20
$begingroup$
Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
$endgroup$
– Nick Peterson
Jan 17 at 21:23
$begingroup$
Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
$endgroup$
– Nick Peterson
Jan 17 at 21:23
$begingroup$
@learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
$endgroup$
– Henry
Jan 17 at 21:24
$begingroup$
@learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
$endgroup$
– Henry
Jan 17 at 21:24
add a comment |
$begingroup$
One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.
answered Jan 17 at 21:33
Peter ForemanPeter Foreman
2,399214
$endgroup$
add a comment |
$begingroup$
One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.
answered Jan 17 at 21:33
Peter ForemanPeter Foreman
2,399214
$endgroup$
add a comment |
$begingroup$
One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.
answered Jan 17 at 21:33
Peter ForemanPeter Foreman
2,399214
$endgroup$
One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.
answered Jan 17 at 21:33
Peter ForemanPeter Foreman
2,399214
answered Jan 17 at 21:33
Peter ForemanPeter Foreman
2,399214
answered Jan 17 at 21:33
Peter ForemanPeter Foreman
2,399214
answered Jan 17 at 21:33
Peter ForemanPeter Foreman
2,399214
2,399214
add a comment |
add a comment |
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