Find $lim_n rightarrowinfty P_n$whreas $P_n=frac2^3-12^3+1cdotfrac3^3-13^3+1cdotcdotcdotfracn^3-1n^3+1$.How to compute $prodlimits^infty_n=2 fracn^3-1n^3+1$$lim_n toinfty p_n = p$ implies $lim_n to inftyp_n^3 = p^3$Looking for fast text-booksHow to find $lim_n to infty int_0^1 cdots int_0^1 sqrtx_1+sqrtx_2+sqrtdots+sqrtx_ndx_1 dx_2dots dx_n$Find $lim_x rightarrow infty(fracxx^2+1cdot e^x)$Show that $lim_ntoinftyb_n=fracsqrtb^2-a^2arccosfracab$$lim_n rightarrow infty a_n = +infty, lim_n rightarrow infty b_n = +infty$ and $lim_n rightarrow infty(a_n + b_n ) = -infty$.Find $lim_nrightarrow inftyfrac(2n-1)!!(2n)!!.$Evaluate: $lim_xto 2fracx-2x^2-4$ without using L'Hôpital RuleEvaluate $lim_x to 0+ (e^1/sin x-e^(1/x))$Finding $lim_n rightarrow inftya_n$ given $lim_n rightarrow inftyfraca_n -1a_n + 1$
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Find $lim_n rightarrowinfty P_n$whreas $P_n=frac2^3-12^3+1cdotfrac3^3-13^3+1cdotcdotcdotfracn^3-1n^3+1$.
How to compute $prodlimits^infty_n=2 fracn^3-1n^3+1$$lim_n toinfty p_n = p$ implies $lim_n to inftyp_n^3 = p^3$Looking for fast text-booksHow to find $lim_n to infty int_0^1 cdots int_0^1 sqrtx_1+sqrtx_2+sqrtdots+sqrtx_ndx_1 dx_2dots dx_n$Find $lim_x rightarrow infty(fracxx^2+1cdot e^x)$Show that $lim_ntoinftyb_n=fracsqrtb^2-a^2arccosfracab$$lim_n rightarrow infty a_n = +infty, lim_n rightarrow infty b_n = +infty$ and $lim_n rightarrow infty(a_n + b_n ) = -infty$.Find $lim_nrightarrow inftyfrac(2n-1)!!(2n)!!.$Evaluate: $lim_xto 2fracx-2x^2-4$ without using L'Hôpital RuleEvaluate $lim_x to 0+ (e^1/sin x-e^(1/x))$Finding $lim_n rightarrow inftya_n$ given $lim_n rightarrow inftyfraca_n -1a_n + 1$
$begingroup$
$lim_n rightarrowinfty P_n$whreas $P_n=frac2^3-12^3+1cdotfrac3^3-13^3+1cdotcdotcdotfracn^3-1n^3+1$.
This is a past problem of a high-school level math compettion.
My tries:
1.Initially I thought of finding the product of the sequence.But then I realize there is no need of product coz' it'll make the function more complicated.
2.Then I tried to simplify first few terms so that may be in some way some of the middle terms may be cancelled out .
So the product of first few terms look sth like this,
$frac79cdotfrac2628cdotfrac6365cdotcdotfracn^3-1n^3+1$
But I could'nt find any pattern in this.So how the limit can be evaluated?
real-analysis limits
asked Mar 29 at 3:41
InvntoInvnto
727
$endgroup$
add a comment |
$begingroup$
$lim_n rightarrowinfty P_n$whreas $P_n=frac2^3-12^3+1cdotfrac3^3-13^3+1cdotcdotcdotfracn^3-1n^3+1$.
This is a past problem of a high-school level math compettion.
My tries:
1.Initially I thought of finding the product of the sequence.But then I realize there is no need of product coz' it'll make the function more complicated.
2.Then I tried to simplify first few terms so that may be in some way some of the middle terms may be cancelled out .
So the product of first few terms look sth like this,
$frac79cdotfrac2628cdotfrac6365cdotcdotfracn^3-1n^3+1$
But I could'nt find any pattern in this.So how the limit can be evaluated?
real-analysis limits
asked Mar 29 at 3:41
InvntoInvnto
727
$endgroup$
$begingroup$
math.stackexchange.com/questions/462082/…
$endgroup$
– lab bhattacharjee
Mar 29 at 3:45
$begingroup$
$displaystyle 2/3$ is the solution (explicitly).
$endgroup$
– Felix Marin
Mar 31 at 2:48
add a comment |
$begingroup$
$lim_n rightarrowinfty P_n$whreas $P_n=frac2^3-12^3+1cdotfrac3^3-13^3+1cdotcdotcdotfracn^3-1n^3+1$.
This is a past problem of a high-school level math compettion.
My tries:
1.Initially I thought of finding the product of the sequence.But then I realize there is no need of product coz' it'll make the function more complicated.
2.Then I tried to simplify first few terms so that may be in some way some of the middle terms may be cancelled out .
So the product of first few terms look sth like this,
$frac79cdotfrac2628cdotfrac6365cdotcdotfracn^3-1n^3+1$
But I could'nt find any pattern in this.So how the limit can be evaluated?
real-analysis limits
asked Mar 29 at 3:41
InvntoInvnto
727
$endgroup$
$lim_n rightarrowinfty P_n$whreas $P_n=frac2^3-12^3+1cdotfrac3^3-13^3+1cdotcdotcdotfracn^3-1n^3+1$.
This is a past problem of a high-school level math compettion.
My tries:
1.Initially I thought of finding the product of the sequence.But then I realize there is no need of product coz' it'll make the function more complicated.
2.Then I tried to simplify first few terms so that may be in some way some of the middle terms may be cancelled out .
So the product of first few terms look sth like this,
$frac79cdotfrac2628cdotfrac6365cdotcdotfracn^3-1n^3+1$
But I could'nt find any pattern in this.So how the limit can be evaluated?
real-analysis limits
real-analysis limits
asked Mar 29 at 3:41
InvntoInvnto
727
asked Mar 29 at 3:41
InvntoInvnto
727
asked Mar 29 at 3:41
InvntoInvnto
727
asked Mar 29 at 3:41
InvntoInvnto
727
asked Mar 29 at 3:41
InvntoInvnto
727
727
$begingroup$
math.stackexchange.com/questions/462082/…
$endgroup$
– lab bhattacharjee
Mar 29 at 3:45
$begingroup$
$displaystyle 2/3$ is the solution (explicitly).
$endgroup$
– Felix Marin
Mar 31 at 2:48
add a comment |
$begingroup$
math.stackexchange.com/questions/462082/…
$endgroup$
– lab bhattacharjee
Mar 29 at 3:45
$begingroup$
$displaystyle 2/3$ is the solution (explicitly).
$endgroup$
– Felix Marin
Mar 31 at 2:48
$begingroup$
math.stackexchange.com/questions/462082/…
$endgroup$
– lab bhattacharjee
Mar 29 at 3:45
$begingroup$
math.stackexchange.com/questions/462082/…
$endgroup$
– lab bhattacharjee
Mar 29 at 3:45
$begingroup$
$displaystyle 2/3$ is the solution (explicitly).
$endgroup$
– Felix Marin
Mar 31 at 2:48
$begingroup$
$displaystyle 2/3$ is the solution (explicitly).
$endgroup$
– Felix Marin
Mar 31 at 2:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There's some telescoping going on here. We have
$$fracn^3-1n^3+1=fracn-1n+1fracn^2+n+1n^2-n+1.$$
Then
$$prod_n=2^Nfracn-1n+1=frac13frac24frac35frac46cdotsfracN-1N+2$$
and
$$prod_n=2^Nfracn^2+n+1n^2-n+1=frac73frac137frac2113frac3121cdotsfracN^2+N+1N^2-N+1.$$
Both these products telescope to give a closed form for
$$prod_n=2^Nfracn^3-1n^3+1.$$
answered Mar 29 at 3:50
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
But I have no idea about the telescoping !!Can't be there some other simple ways!!
$endgroup$
– Invnto
Mar 29 at 4:06
1
$begingroup$
@Invnto he actually gave the answer. Look more closely.
$endgroup$
– amitava
Mar 29 at 4:10
$begingroup$
yeah!! Now I got it !!Actually I haven't observed the solution properly!!
$endgroup$
– Invnto
Mar 29 at 4:19
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
bbox[10px,#ffd]P_n & equiv
prod_k = 2^nk^3 - 1 over k^3 + 1 =
prod_k = 2^nparsk - 1parsk - expo2piic/3
parsk - expo-2piic/3 over
parsk + 1parsk - expopiic/3
parsk - expo-piic/3
\[5mm] & =
parsn - 1! over parsn + 1!/2prod_k = 2^nparsk - expo2piic/3
parsk - expo-2piic/3 over
parsk - expopiic/3parsk - expo-piic/3
\[5mm] & =
2 over parsn + 1nvertsprod_k = 2^n
parsk - expo2piic/3 over parsk - expopiic/3^2
\[5mm] & =
2 over nparsn + 1vertsprod_k = 2^n
parsk + 1/2 - root3ic/2 over
parsk - 1/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1
vertspars5/2 - root3ic/2^overlinen - 1 over
pars3/2 - root3ic/2^overlinen - 1^2
\[5mm] & =
2 over nparsn + 1
vertsGammaparsn + 3/2 - root3ic/2/Gammapars5/2 - root3ic/2 over
Gammaparsn + 1/2 - root3ic/2/Gammapars3/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1,vertsGammapars3/2 - root3ic/2 over Gammapars5/2 - root3ic/2^2
vertsparsn + 1/2 - root3ic/2! over
parsn - 1/2 - root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2 over nparsn + 1
overbrace1 over verts3/2 - root3ic/2^2^ds= 1/3
times
\[2mm] &
vertsroot2piparsn + 1/2 - root3ic/2^n + 1 - root3ic/2 expo-n - 1/2 + root3ic/2 over
root2piparsn - 1/2 - root3ic/2^n - root3ic/2
expo-n + 1/2 + root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2/3 over nparsn + 1
vertsn^n + 1 - root3ic/2bracks1 + pars1/2 - root3ic/2/n^n expo- 1/2 + root3ic/2 over
n^n - root3ic/2bracks1 - pars1/2 + root3ic/2/n^n
expo1/2 + root3ic/2^2
\[5mm] & =
2/3 over nparsn + 1,n^2
,,,stackrelmrmas n to inftyLargeto,,,
bbx2 over 3
endalign
answered Mar 30 at 6:43
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$begingroup$
Not sure this is the method intended for a high school competition :P
$endgroup$
– YiFan
Mar 31 at 2:57
$begingroup$
@YiFan Unfortunately, I agree with you. Anyway, it's an alternative. At least I show that the answer is $displaystyle 2/3$. Thanks.
$endgroup$
– Felix Marin
Mar 31 at 3:01
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There's some telescoping going on here. We have
$$fracn^3-1n^3+1=fracn-1n+1fracn^2+n+1n^2-n+1.$$
Then
$$prod_n=2^Nfracn-1n+1=frac13frac24frac35frac46cdotsfracN-1N+2$$
and
$$prod_n=2^Nfracn^2+n+1n^2-n+1=frac73frac137frac2113frac3121cdotsfracN^2+N+1N^2-N+1.$$
Both these products telescope to give a closed form for
$$prod_n=2^Nfracn^3-1n^3+1.$$
answered Mar 29 at 3:50
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
But I have no idea about the telescoping !!Can't be there some other simple ways!!
$endgroup$
– Invnto
Mar 29 at 4:06
1
$begingroup$
@Invnto he actually gave the answer. Look more closely.
$endgroup$
– amitava
Mar 29 at 4:10
$begingroup$
yeah!! Now I got it !!Actually I haven't observed the solution properly!!
$endgroup$
– Invnto
Mar 29 at 4:19
add a comment |
$begingroup$
There's some telescoping going on here. We have
$$fracn^3-1n^3+1=fracn-1n+1fracn^2+n+1n^2-n+1.$$
Then
$$prod_n=2^Nfracn-1n+1=frac13frac24frac35frac46cdotsfracN-1N+2$$
and
$$prod_n=2^Nfracn^2+n+1n^2-n+1=frac73frac137frac2113frac3121cdotsfracN^2+N+1N^2-N+1.$$
Both these products telescope to give a closed form for
$$prod_n=2^Nfracn^3-1n^3+1.$$
answered Mar 29 at 3:50
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
But I have no idea about the telescoping !!Can't be there some other simple ways!!
$endgroup$
– Invnto
Mar 29 at 4:06
1
$begingroup$
@Invnto he actually gave the answer. Look more closely.
$endgroup$
– amitava
Mar 29 at 4:10
$begingroup$
yeah!! Now I got it !!Actually I haven't observed the solution properly!!
$endgroup$
– Invnto
Mar 29 at 4:19
add a comment |
$begingroup$
There's some telescoping going on here. We have
$$fracn^3-1n^3+1=fracn-1n+1fracn^2+n+1n^2-n+1.$$
Then
$$prod_n=2^Nfracn-1n+1=frac13frac24frac35frac46cdotsfracN-1N+2$$
and
$$prod_n=2^Nfracn^2+n+1n^2-n+1=frac73frac137frac2113frac3121cdotsfracN^2+N+1N^2-N+1.$$
Both these products telescope to give a closed form for
$$prod_n=2^Nfracn^3-1n^3+1.$$
answered Mar 29 at 3:50
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
There's some telescoping going on here. We have
$$fracn^3-1n^3+1=fracn-1n+1fracn^2+n+1n^2-n+1.$$
Then
$$prod_n=2^Nfracn-1n+1=frac13frac24frac35frac46cdotsfracN-1N+2$$
and
$$prod_n=2^Nfracn^2+n+1n^2-n+1=frac73frac137frac2113frac3121cdotsfracN^2+N+1N^2-N+1.$$
Both these products telescope to give a closed form for
$$prod_n=2^Nfracn^3-1n^3+1.$$
answered Mar 29 at 3:50
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 29 at 3:50
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 29 at 3:50
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 29 at 3:50
Lord Shark the UnknownLord Shark the Unknown
108k1162135
108k1162135
$begingroup$
But I have no idea about the telescoping !!Can't be there some other simple ways!!
$endgroup$
– Invnto
Mar 29 at 4:06
1
$begingroup$
@Invnto he actually gave the answer. Look more closely.
$endgroup$
– amitava
Mar 29 at 4:10
$begingroup$
yeah!! Now I got it !!Actually I haven't observed the solution properly!!
$endgroup$
– Invnto
Mar 29 at 4:19
add a comment |
$begingroup$
But I have no idea about the telescoping !!Can't be there some other simple ways!!
$endgroup$
– Invnto
Mar 29 at 4:06
1
$begingroup$
@Invnto he actually gave the answer. Look more closely.
$endgroup$
– amitava
Mar 29 at 4:10
$begingroup$
yeah!! Now I got it !!Actually I haven't observed the solution properly!!
$endgroup$
– Invnto
Mar 29 at 4:19
$begingroup$
But I have no idea about the telescoping !!Can't be there some other simple ways!!
$endgroup$
– Invnto
Mar 29 at 4:06
$begingroup$
But I have no idea about the telescoping !!Can't be there some other simple ways!!
$endgroup$
– Invnto
Mar 29 at 4:06
1
1
$begingroup$
@Invnto he actually gave the answer. Look more closely.
$endgroup$
– amitava
Mar 29 at 4:10
$begingroup$
@Invnto he actually gave the answer. Look more closely.
$endgroup$
– amitava
Mar 29 at 4:10
$begingroup$
yeah!! Now I got it !!Actually I haven't observed the solution properly!!
$endgroup$
– Invnto
Mar 29 at 4:19
$begingroup$
yeah!! Now I got it !!Actually I haven't observed the solution properly!!
$endgroup$
– Invnto
Mar 29 at 4:19
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
bbox[10px,#ffd]P_n & equiv
prod_k = 2^nk^3 - 1 over k^3 + 1 =
prod_k = 2^nparsk - 1parsk - expo2piic/3
parsk - expo-2piic/3 over
parsk + 1parsk - expopiic/3
parsk - expo-piic/3
\[5mm] & =
parsn - 1! over parsn + 1!/2prod_k = 2^nparsk - expo2piic/3
parsk - expo-2piic/3 over
parsk - expopiic/3parsk - expo-piic/3
\[5mm] & =
2 over parsn + 1nvertsprod_k = 2^n
parsk - expo2piic/3 over parsk - expopiic/3^2
\[5mm] & =
2 over nparsn + 1vertsprod_k = 2^n
parsk + 1/2 - root3ic/2 over
parsk - 1/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1
vertspars5/2 - root3ic/2^overlinen - 1 over
pars3/2 - root3ic/2^overlinen - 1^2
\[5mm] & =
2 over nparsn + 1
vertsGammaparsn + 3/2 - root3ic/2/Gammapars5/2 - root3ic/2 over
Gammaparsn + 1/2 - root3ic/2/Gammapars3/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1,vertsGammapars3/2 - root3ic/2 over Gammapars5/2 - root3ic/2^2
vertsparsn + 1/2 - root3ic/2! over
parsn - 1/2 - root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2 over nparsn + 1
overbrace1 over verts3/2 - root3ic/2^2^ds= 1/3
times
\[2mm] &
vertsroot2piparsn + 1/2 - root3ic/2^n + 1 - root3ic/2 expo-n - 1/2 + root3ic/2 over
root2piparsn - 1/2 - root3ic/2^n - root3ic/2
expo-n + 1/2 + root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2/3 over nparsn + 1
vertsn^n + 1 - root3ic/2bracks1 + pars1/2 - root3ic/2/n^n expo- 1/2 + root3ic/2 over
n^n - root3ic/2bracks1 - pars1/2 + root3ic/2/n^n
expo1/2 + root3ic/2^2
\[5mm] & =
2/3 over nparsn + 1,n^2
,,,stackrelmrmas n to inftyLargeto,,,
bbx2 over 3
endalign
answered Mar 30 at 6:43
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$begingroup$
Not sure this is the method intended for a high school competition :P
$endgroup$
– YiFan
Mar 31 at 2:57
$begingroup$
@YiFan Unfortunately, I agree with you. Anyway, it's an alternative. At least I show that the answer is $displaystyle 2/3$. Thanks.
$endgroup$
– Felix Marin
Mar 31 at 3:01
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
bbox[10px,#ffd]P_n & equiv
prod_k = 2^nk^3 - 1 over k^3 + 1 =
prod_k = 2^nparsk - 1parsk - expo2piic/3
parsk - expo-2piic/3 over
parsk + 1parsk - expopiic/3
parsk - expo-piic/3
\[5mm] & =
parsn - 1! over parsn + 1!/2prod_k = 2^nparsk - expo2piic/3
parsk - expo-2piic/3 over
parsk - expopiic/3parsk - expo-piic/3
\[5mm] & =
2 over parsn + 1nvertsprod_k = 2^n
parsk - expo2piic/3 over parsk - expopiic/3^2
\[5mm] & =
2 over nparsn + 1vertsprod_k = 2^n
parsk + 1/2 - root3ic/2 over
parsk - 1/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1
vertspars5/2 - root3ic/2^overlinen - 1 over
pars3/2 - root3ic/2^overlinen - 1^2
\[5mm] & =
2 over nparsn + 1
vertsGammaparsn + 3/2 - root3ic/2/Gammapars5/2 - root3ic/2 over
Gammaparsn + 1/2 - root3ic/2/Gammapars3/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1,vertsGammapars3/2 - root3ic/2 over Gammapars5/2 - root3ic/2^2
vertsparsn + 1/2 - root3ic/2! over
parsn - 1/2 - root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2 over nparsn + 1
overbrace1 over verts3/2 - root3ic/2^2^ds= 1/3
times
\[2mm] &
vertsroot2piparsn + 1/2 - root3ic/2^n + 1 - root3ic/2 expo-n - 1/2 + root3ic/2 over
root2piparsn - 1/2 - root3ic/2^n - root3ic/2
expo-n + 1/2 + root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2/3 over nparsn + 1
vertsn^n + 1 - root3ic/2bracks1 + pars1/2 - root3ic/2/n^n expo- 1/2 + root3ic/2 over
n^n - root3ic/2bracks1 - pars1/2 + root3ic/2/n^n
expo1/2 + root3ic/2^2
\[5mm] & =
2/3 over nparsn + 1,n^2
,,,stackrelmrmas n to inftyLargeto,,,
bbx2 over 3
endalign
answered Mar 30 at 6:43
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$begingroup$
Not sure this is the method intended for a high school competition :P
$endgroup$
– YiFan
Mar 31 at 2:57
$begingroup$
@YiFan Unfortunately, I agree with you. Anyway, it's an alternative. At least I show that the answer is $displaystyle 2/3$. Thanks.
$endgroup$
– Felix Marin
Mar 31 at 3:01
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
bbox[10px,#ffd]P_n & equiv
prod_k = 2^nk^3 - 1 over k^3 + 1 =
prod_k = 2^nparsk - 1parsk - expo2piic/3
parsk - expo-2piic/3 over
parsk + 1parsk - expopiic/3
parsk - expo-piic/3
\[5mm] & =
parsn - 1! over parsn + 1!/2prod_k = 2^nparsk - expo2piic/3
parsk - expo-2piic/3 over
parsk - expopiic/3parsk - expo-piic/3
\[5mm] & =
2 over parsn + 1nvertsprod_k = 2^n
parsk - expo2piic/3 over parsk - expopiic/3^2
\[5mm] & =
2 over nparsn + 1vertsprod_k = 2^n
parsk + 1/2 - root3ic/2 over
parsk - 1/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1
vertspars5/2 - root3ic/2^overlinen - 1 over
pars3/2 - root3ic/2^overlinen - 1^2
\[5mm] & =
2 over nparsn + 1
vertsGammaparsn + 3/2 - root3ic/2/Gammapars5/2 - root3ic/2 over
Gammaparsn + 1/2 - root3ic/2/Gammapars3/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1,vertsGammapars3/2 - root3ic/2 over Gammapars5/2 - root3ic/2^2
vertsparsn + 1/2 - root3ic/2! over
parsn - 1/2 - root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2 over nparsn + 1
overbrace1 over verts3/2 - root3ic/2^2^ds= 1/3
times
\[2mm] &
vertsroot2piparsn + 1/2 - root3ic/2^n + 1 - root3ic/2 expo-n - 1/2 + root3ic/2 over
root2piparsn - 1/2 - root3ic/2^n - root3ic/2
expo-n + 1/2 + root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2/3 over nparsn + 1
vertsn^n + 1 - root3ic/2bracks1 + pars1/2 - root3ic/2/n^n expo- 1/2 + root3ic/2 over
n^n - root3ic/2bracks1 - pars1/2 + root3ic/2/n^n
expo1/2 + root3ic/2^2
\[5mm] & =
2/3 over nparsn + 1,n^2
,,,stackrelmrmas n to inftyLargeto,,,
bbx2 over 3
endalign
answered Mar 30 at 6:43
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
bbox[10px,#ffd]P_n & equiv
prod_k = 2^nk^3 - 1 over k^3 + 1 =
prod_k = 2^nparsk - 1parsk - expo2piic/3
parsk - expo-2piic/3 over
parsk + 1parsk - expopiic/3
parsk - expo-piic/3
\[5mm] & =
parsn - 1! over parsn + 1!/2prod_k = 2^nparsk - expo2piic/3
parsk - expo-2piic/3 over
parsk - expopiic/3parsk - expo-piic/3
\[5mm] & =
2 over parsn + 1nvertsprod_k = 2^n
parsk - expo2piic/3 over parsk - expopiic/3^2
\[5mm] & =
2 over nparsn + 1vertsprod_k = 2^n
parsk + 1/2 - root3ic/2 over
parsk - 1/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1
vertspars5/2 - root3ic/2^overlinen - 1 over
pars3/2 - root3ic/2^overlinen - 1^2
\[5mm] & =
2 over nparsn + 1
vertsGammaparsn + 3/2 - root3ic/2/Gammapars5/2 - root3ic/2 over
Gammaparsn + 1/2 - root3ic/2/Gammapars3/2 - root3ic/2^2
\[5mm] & =
2 over nparsn + 1,vertsGammapars3/2 - root3ic/2 over Gammapars5/2 - root3ic/2^2
vertsparsn + 1/2 - root3ic/2! over
parsn - 1/2 - root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2 over nparsn + 1
overbrace1 over verts3/2 - root3ic/2^2^ds= 1/3
times
\[2mm] &
vertsroot2piparsn + 1/2 - root3ic/2^n + 1 - root3ic/2 expo-n - 1/2 + root3ic/2 over
root2piparsn - 1/2 - root3ic/2^n - root3ic/2
expo-n + 1/2 + root3ic/2^2
\[8mm] & stackrelmrmas n to inftysim,,,
2/3 over nparsn + 1
vertsn^n + 1 - root3ic/2bracks1 + pars1/2 - root3ic/2/n^n expo- 1/2 + root3ic/2 over
n^n - root3ic/2bracks1 - pars1/2 + root3ic/2/n^n
expo1/2 + root3ic/2^2
\[5mm] & =
2/3 over nparsn + 1,n^2
,,,stackrelmrmas n to inftyLargeto,,,
bbx2 over 3
endalign
answered Mar 30 at 6:43
Felix MarinFelix Marin
68.8k7109146
edited Mar 31 at 2:44
answered Mar 30 at 6:43
Felix MarinFelix Marin
68.8k7109146
answered Mar 30 at 6:43
Felix MarinFelix Marin
68.8k7109146
answered Mar 30 at 6:43
Felix MarinFelix Marin
68.8k7109146
68.8k7109146
$begingroup$
Not sure this is the method intended for a high school competition :P
$endgroup$
– YiFan
Mar 31 at 2:57
$begingroup$
@YiFan Unfortunately, I agree with you. Anyway, it's an alternative. At least I show that the answer is $displaystyle 2/3$. Thanks.
$endgroup$
– Felix Marin
Mar 31 at 3:01
add a comment |
$begingroup$
Not sure this is the method intended for a high school competition :P
$endgroup$
– YiFan
Mar 31 at 2:57
$begingroup$
@YiFan Unfortunately, I agree with you. Anyway, it's an alternative. At least I show that the answer is $displaystyle 2/3$. Thanks.
$endgroup$
– Felix Marin
Mar 31 at 3:01
$begingroup$
Not sure this is the method intended for a high school competition :P
$endgroup$
– YiFan
Mar 31 at 2:57
$begingroup$
Not sure this is the method intended for a high school competition :P
$endgroup$
– YiFan
Mar 31 at 2:57
$begingroup$
@YiFan Unfortunately, I agree with you. Anyway, it's an alternative. At least I show that the answer is $displaystyle 2/3$. Thanks.
$endgroup$
– Felix Marin
Mar 31 at 3:01
$begingroup$
@YiFan Unfortunately, I agree with you. Anyway, it's an alternative. At least I show that the answer is $displaystyle 2/3$. Thanks.
$endgroup$
– Felix Marin
Mar 31 at 3:01
add a comment |
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$begingroup$
math.stackexchange.com/questions/462082/…
$endgroup$
– lab bhattacharjee
Mar 29 at 3:45
$begingroup$
$displaystyle 2/3$ is the solution (explicitly).
$endgroup$
– Felix Marin
Mar 31 at 2:48