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Find $t$ such that $lim_ntoinfty frac sum_r=1^n r^4cdotsum_r=1^n r^5sum_r=1^n r^tcdotsum_r=1^n r^9-t=frac 45$
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Find $t$ such that $$lim_ntoinfty frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)=frac 45.$$
At first sight this question scared the hell out of me. I tried using the general known formulas like $$sum_r=1^n r^4=frac n(n+1)(2n+1)(3n^2+3n-1)6$$ and $$sum_r=1^n r^5=frac n^2(n+1)^2(2n^2+2n-1)12.$$
But the denominator portion really doesn't go with it. I tried to write it in form of integrals. I also searched the internet for some information but it dealt higher level calculus relating the harmonic functions, Bernoulli numbers and the Riemann zeta function. I read about it but couldn't get much out of it.
Any help would be greatly appreciated. Thanks.
calculus sequences-and-series limits
asked Jul 20 at 15:38
Manthanein
6,1291437
add a comment |Â
up vote
3
down vote
favorite
Find $t$ such that $$lim_ntoinfty frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)=frac 45.$$
At first sight this question scared the hell out of me. I tried using the general known formulas like $$sum_r=1^n r^4=frac n(n+1)(2n+1)(3n^2+3n-1)6$$ and $$sum_r=1^n r^5=frac n^2(n+1)^2(2n^2+2n-1)12.$$
But the denominator portion really doesn't go with it. I tried to write it in form of integrals. I also searched the internet for some information but it dealt higher level calculus relating the harmonic functions, Bernoulli numbers and the Riemann zeta function. I read about it but couldn't get much out of it.
Any help would be greatly appreciated. Thanks.
calculus sequences-and-series limits
asked Jul 20 at 15:38
Manthanein
6,1291437
2
Use $sum_1^n r^tsim n^t+1/(t+1).$
– Lord Shark the Unknown
Jul 20 at 15:42
You probably just need to prove $sum_k=1^n k^r sim c_r n^r+1$ for some $c_r$ that does not depend on $n$.
– Arin Chaudhuri
Jul 20 at 15:43
Use math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Jul 20 at 15:44
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find $t$ such that $$lim_ntoinfty frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)=frac 45.$$
At first sight this question scared the hell out of me. I tried using the general known formulas like $$sum_r=1^n r^4=frac n(n+1)(2n+1)(3n^2+3n-1)6$$ and $$sum_r=1^n r^5=frac n^2(n+1)^2(2n^2+2n-1)12.$$
But the denominator portion really doesn't go with it. I tried to write it in form of integrals. I also searched the internet for some information but it dealt higher level calculus relating the harmonic functions, Bernoulli numbers and the Riemann zeta function. I read about it but couldn't get much out of it.
Any help would be greatly appreciated. Thanks.
calculus sequences-and-series limits
asked Jul 20 at 15:38
Manthanein
6,1291437
Find $t$ such that $$lim_ntoinfty frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)=frac 45.$$
At first sight this question scared the hell out of me. I tried using the general known formulas like $$sum_r=1^n r^4=frac n(n+1)(2n+1)(3n^2+3n-1)6$$ and $$sum_r=1^n r^5=frac n^2(n+1)^2(2n^2+2n-1)12.$$
But the denominator portion really doesn't go with it. I tried to write it in form of integrals. I also searched the internet for some information but it dealt higher level calculus relating the harmonic functions, Bernoulli numbers and the Riemann zeta function. I read about it but couldn't get much out of it.
Any help would be greatly appreciated. Thanks.
calculus sequences-and-series limits
asked Jul 20 at 15:38
Manthanein
6,1291437
edited Jul 20 at 15:54
asked Jul 20 at 15:38
Manthanein
6,1291437
asked Jul 20 at 15:38
Manthanein
6,1291437
asked Jul 20 at 15:38
Manthanein
6,1291437
6,1291437
2
Use $sum_1^n r^tsim n^t+1/(t+1).$
– Lord Shark the Unknown
Jul 20 at 15:42
You probably just need to prove $sum_k=1^n k^r sim c_r n^r+1$ for some $c_r$ that does not depend on $n$.
– Arin Chaudhuri
Jul 20 at 15:43
Use math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Jul 20 at 15:44
add a comment |Â
2
Use $sum_1^n r^tsim n^t+1/(t+1).$
– Lord Shark the Unknown
Jul 20 at 15:42
You probably just need to prove $sum_k=1^n k^r sim c_r n^r+1$ for some $c_r$ that does not depend on $n$.
– Arin Chaudhuri
Jul 20 at 15:43
Use math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Jul 20 at 15:44
2
2
Use $sum_1^n r^tsim n^t+1/(t+1).$
– Lord Shark the Unknown
Jul 20 at 15:42
Use $sum_1^n r^tsim n^t+1/(t+1).$
– Lord Shark the Unknown
Jul 20 at 15:42
You probably just need to prove $sum_k=1^n k^r sim c_r n^r+1$ for some $c_r$ that does not depend on $n$.
– Arin Chaudhuri
Jul 20 at 15:43
You probably just need to prove $sum_k=1^n k^r sim c_r n^r+1$ for some $c_r$ that does not depend on $n$.
– Arin Chaudhuri
Jul 20 at 15:43
Use math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Jul 20 at 15:44
Use math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Jul 20 at 15:44
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Hint. Note that for $a>0$
$$F(a):=lim_ntoinfty frac1n^a+1left(sum_r=1^n r^aright)=
lim_ntoinfty frac1nleft(sum_r=1^n left(fracrnright)^aright)to int_0^1x^a dx=frac1a+1.$$
Then, as $ntoinfty$,
$$frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)
=fracfrac1n^5left(sum_r=1^n r^4right)cdotfrac1n^6left(sum_r=1^n r^5right)frac1n^t+1left(sum_r=1^n r^tright)cdotfrac1n^10-tleft(sum_r=1^n r^9-tright)
tofracF(4)cdot F(5)F(t)cdot F(9-t).$$
answered Jul 20 at 15:44
Robert Z
84k954122
Oh yes I should have thought of Riemann sums and integrals
– Manthanein
Jul 20 at 15:53
add a comment |Â
up vote
2
down vote
The skeleton:
First thought is to apply integrals:
$$
dfracintlimits_1^n x^4 dx intlimits_1^n x^5 dxintlimits_1^n x^t dx intlimits_1^n x^9-t dx sim
dfracdfracn^55 cdot dfracn^66dfracn^t+1t+1 cdot dfracn^9-t+19-t+1
$$
Then we'll have fraction
$$
dfrac(9-t+1)(t+1)30=dfrac45 = dfrac2430
$$
Then quadratic equation
$$(9-t+1)(t+1) = 24$$
So $t=2$, $t=7$ should work.
answered Jul 20 at 15:49
Oleg567
13.6k22969
add a comment |Â
up vote
0
down vote
The key here is Cesaro-Stolz. Using Cesaro-Stolz we have $$f(t) =lim_ntoinfty frac1n^t+1sum_r=1^nr^t=lim_nto infty fracn^t n^t+1-(n-1)^t+1=frac1t+1$$ Dividing the numerator and denominator of the given expression by $n^11$ we can see that the desired limit is $$fracf(4)f(5)f(t)f(9-t)$$ and now you can equate this to $4/5$ and get $t$.
answered Jul 21 at 2:42
Paramanand Singh
45.2k553142
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint. Note that for $a>0$
$$F(a):=lim_ntoinfty frac1n^a+1left(sum_r=1^n r^aright)=
lim_ntoinfty frac1nleft(sum_r=1^n left(fracrnright)^aright)to int_0^1x^a dx=frac1a+1.$$
Then, as $ntoinfty$,
$$frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)
=fracfrac1n^5left(sum_r=1^n r^4right)cdotfrac1n^6left(sum_r=1^n r^5right)frac1n^t+1left(sum_r=1^n r^tright)cdotfrac1n^10-tleft(sum_r=1^n r^9-tright)
tofracF(4)cdot F(5)F(t)cdot F(9-t).$$
answered Jul 20 at 15:44
Robert Z
84k954122
Oh yes I should have thought of Riemann sums and integrals
– Manthanein
Jul 20 at 15:53
add a comment |Â
up vote
4
down vote
accepted
Hint. Note that for $a>0$
$$F(a):=lim_ntoinfty frac1n^a+1left(sum_r=1^n r^aright)=
lim_ntoinfty frac1nleft(sum_r=1^n left(fracrnright)^aright)to int_0^1x^a dx=frac1a+1.$$
Then, as $ntoinfty$,
$$frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)
=fracfrac1n^5left(sum_r=1^n r^4right)cdotfrac1n^6left(sum_r=1^n r^5right)frac1n^t+1left(sum_r=1^n r^tright)cdotfrac1n^10-tleft(sum_r=1^n r^9-tright)
tofracF(4)cdot F(5)F(t)cdot F(9-t).$$
answered Jul 20 at 15:44
Robert Z
84k954122
Oh yes I should have thought of Riemann sums and integrals
– Manthanein
Jul 20 at 15:53
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint. Note that for $a>0$
$$F(a):=lim_ntoinfty frac1n^a+1left(sum_r=1^n r^aright)=
lim_ntoinfty frac1nleft(sum_r=1^n left(fracrnright)^aright)to int_0^1x^a dx=frac1a+1.$$
Then, as $ntoinfty$,
$$frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)
=fracfrac1n^5left(sum_r=1^n r^4right)cdotfrac1n^6left(sum_r=1^n r^5right)frac1n^t+1left(sum_r=1^n r^tright)cdotfrac1n^10-tleft(sum_r=1^n r^9-tright)
tofracF(4)cdot F(5)F(t)cdot F(9-t).$$
answered Jul 20 at 15:44
Robert Z
84k954122
Hint. Note that for $a>0$
$$F(a):=lim_ntoinfty frac1n^a+1left(sum_r=1^n r^aright)=
lim_ntoinfty frac1nleft(sum_r=1^n left(fracrnright)^aright)to int_0^1x^a dx=frac1a+1.$$
Then, as $ntoinfty$,
$$frac left(sum_r=1^n r^4right)cdotleft(sum_r=1^n r^5right)left(sum_r=1^n r^tright)cdotleft(sum_r=1^n r^9-tright)
=fracfrac1n^5left(sum_r=1^n r^4right)cdotfrac1n^6left(sum_r=1^n r^5right)frac1n^t+1left(sum_r=1^n r^tright)cdotfrac1n^10-tleft(sum_r=1^n r^9-tright)
tofracF(4)cdot F(5)F(t)cdot F(9-t).$$
answered Jul 20 at 15:44
Robert Z
84k954122
edited Jul 20 at 15:57
answered Jul 20 at 15:44
Robert Z
84k954122
answered Jul 20 at 15:44
Robert Z
84k954122
answered Jul 20 at 15:44
Robert Z
84k954122
84k954122
Oh yes I should have thought of Riemann sums and integrals
– Manthanein
Jul 20 at 15:53
add a comment |Â
Oh yes I should have thought of Riemann sums and integrals
– Manthanein
Jul 20 at 15:53
Oh yes I should have thought of Riemann sums and integrals
– Manthanein
Jul 20 at 15:53
Oh yes I should have thought of Riemann sums and integrals
– Manthanein
Jul 20 at 15:53
add a comment |Â
up vote
2
down vote
The skeleton:
First thought is to apply integrals:
$$
dfracintlimits_1^n x^4 dx intlimits_1^n x^5 dxintlimits_1^n x^t dx intlimits_1^n x^9-t dx sim
dfracdfracn^55 cdot dfracn^66dfracn^t+1t+1 cdot dfracn^9-t+19-t+1
$$
Then we'll have fraction
$$
dfrac(9-t+1)(t+1)30=dfrac45 = dfrac2430
$$
Then quadratic equation
$$(9-t+1)(t+1) = 24$$
So $t=2$, $t=7$ should work.
answered Jul 20 at 15:49
Oleg567
13.6k22969
add a comment |Â
up vote
2
down vote
The skeleton:
First thought is to apply integrals:
$$
dfracintlimits_1^n x^4 dx intlimits_1^n x^5 dxintlimits_1^n x^t dx intlimits_1^n x^9-t dx sim
dfracdfracn^55 cdot dfracn^66dfracn^t+1t+1 cdot dfracn^9-t+19-t+1
$$
Then we'll have fraction
$$
dfrac(9-t+1)(t+1)30=dfrac45 = dfrac2430
$$
Then quadratic equation
$$(9-t+1)(t+1) = 24$$
So $t=2$, $t=7$ should work.
answered Jul 20 at 15:49
Oleg567
13.6k22969
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The skeleton:
First thought is to apply integrals:
$$
dfracintlimits_1^n x^4 dx intlimits_1^n x^5 dxintlimits_1^n x^t dx intlimits_1^n x^9-t dx sim
dfracdfracn^55 cdot dfracn^66dfracn^t+1t+1 cdot dfracn^9-t+19-t+1
$$
Then we'll have fraction
$$
dfrac(9-t+1)(t+1)30=dfrac45 = dfrac2430
$$
Then quadratic equation
$$(9-t+1)(t+1) = 24$$
So $t=2$, $t=7$ should work.
answered Jul 20 at 15:49
Oleg567
13.6k22969
The skeleton:
First thought is to apply integrals:
$$
dfracintlimits_1^n x^4 dx intlimits_1^n x^5 dxintlimits_1^n x^t dx intlimits_1^n x^9-t dx sim
dfracdfracn^55 cdot dfracn^66dfracn^t+1t+1 cdot dfracn^9-t+19-t+1
$$
Then we'll have fraction
$$
dfrac(9-t+1)(t+1)30=dfrac45 = dfrac2430
$$
Then quadratic equation
$$(9-t+1)(t+1) = 24$$
So $t=2$, $t=7$ should work.
answered Jul 20 at 15:49
Oleg567
13.6k22969
answered Jul 20 at 15:49
Oleg567
13.6k22969
answered Jul 20 at 15:49
Oleg567
13.6k22969
answered Jul 20 at 15:49
Oleg567
13.6k22969
13.6k22969
add a comment |Â
add a comment |Â
up vote
0
down vote
The key here is Cesaro-Stolz. Using Cesaro-Stolz we have $$f(t) =lim_ntoinfty frac1n^t+1sum_r=1^nr^t=lim_nto infty fracn^t n^t+1-(n-1)^t+1=frac1t+1$$ Dividing the numerator and denominator of the given expression by $n^11$ we can see that the desired limit is $$fracf(4)f(5)f(t)f(9-t)$$ and now you can equate this to $4/5$ and get $t$.
answered Jul 21 at 2:42
Paramanand Singh
45.2k553142
add a comment |Â
up vote
0
down vote
The key here is Cesaro-Stolz. Using Cesaro-Stolz we have $$f(t) =lim_ntoinfty frac1n^t+1sum_r=1^nr^t=lim_nto infty fracn^t n^t+1-(n-1)^t+1=frac1t+1$$ Dividing the numerator and denominator of the given expression by $n^11$ we can see that the desired limit is $$fracf(4)f(5)f(t)f(9-t)$$ and now you can equate this to $4/5$ and get $t$.
answered Jul 21 at 2:42
Paramanand Singh
45.2k553142
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The key here is Cesaro-Stolz. Using Cesaro-Stolz we have $$f(t) =lim_ntoinfty frac1n^t+1sum_r=1^nr^t=lim_nto infty fracn^t n^t+1-(n-1)^t+1=frac1t+1$$ Dividing the numerator and denominator of the given expression by $n^11$ we can see that the desired limit is $$fracf(4)f(5)f(t)f(9-t)$$ and now you can equate this to $4/5$ and get $t$.
answered Jul 21 at 2:42
Paramanand Singh
45.2k553142
The key here is Cesaro-Stolz. Using Cesaro-Stolz we have $$f(t) =lim_ntoinfty frac1n^t+1sum_r=1^nr^t=lim_nto infty fracn^t n^t+1-(n-1)^t+1=frac1t+1$$ Dividing the numerator and denominator of the given expression by $n^11$ we can see that the desired limit is $$fracf(4)f(5)f(t)f(9-t)$$ and now you can equate this to $4/5$ and get $t$.
answered Jul 21 at 2:42
Paramanand Singh
45.2k553142
answered Jul 21 at 2:42
Paramanand Singh
45.2k553142
answered Jul 21 at 2:42
Paramanand Singh
45.2k553142
answered Jul 21 at 2:42
Paramanand Singh
45.2k553142
45.2k553142
add a comment |Â
add a comment |Â
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2
Use $sum_1^n r^tsim n^t+1/(t+1).$
– Lord Shark the Unknown
Jul 20 at 15:42
You probably just need to prove $sum_k=1^n k^r sim c_r n^r+1$ for some $c_r$ that does not depend on $n$.
– Arin Chaudhuri
Jul 20 at 15:43
Use math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Jul 20 at 15:44