How to show $lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4$?
How would I show $$lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$
My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(zeta_k)$ and $(1/n)=Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.
I think I should find the formula for $$sum_{k=1}^nk^3.$$ This would be $1+8+27+cdots+n^3$... but then I get stuck.
real-analysis definite-integrals riemann-integration riemann-sum
asked Dec 3 '18 at 14:29
kaisakaisa
1019
add a comment |
How would I show $$lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$
My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(zeta_k)$ and $(1/n)=Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.
I think I should find the formula for $$sum_{k=1}^nk^3.$$ This would be $1+8+27+cdots+n^3$... but then I get stuck.
real-analysis definite-integrals riemann-integration riemann-sum
asked Dec 3 '18 at 14:29
kaisakaisa
1019
Ask Faulhaber...
– Yves Daoust
Dec 3 '18 at 14:30
1
As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
– Yves Daoust
Dec 3 '18 at 14:31
1
This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
– Did
Dec 3 '18 at 14:52
add a comment |
How would I show $$lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$
My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(zeta_k)$ and $(1/n)=Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.
I think I should find the formula for $$sum_{k=1}^nk^3.$$ This would be $1+8+27+cdots+n^3$... but then I get stuck.
real-analysis definite-integrals riemann-integration riemann-sum
asked Dec 3 '18 at 14:29
kaisakaisa
1019
How would I show $$lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$
My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(zeta_k)$ and $(1/n)=Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.
I think I should find the formula for $$sum_{k=1}^nk^3.$$ This would be $1+8+27+cdots+n^3$... but then I get stuck.
real-analysis definite-integrals riemann-integration riemann-sum
real-analysis definite-integrals riemann-integration riemann-sum
asked Dec 3 '18 at 14:29
kaisakaisa
1019
asked Dec 3 '18 at 14:29
kaisakaisa
1019
asked Dec 3 '18 at 14:29
kaisakaisa
1019
asked Dec 3 '18 at 14:29
kaisakaisa
1019
asked Dec 3 '18 at 14:29
kaisakaisa
1019
1019
Ask Faulhaber...
– Yves Daoust
Dec 3 '18 at 14:30
1
As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
– Yves Daoust
Dec 3 '18 at 14:31
1
This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
– Did
Dec 3 '18 at 14:52
add a comment |
Ask Faulhaber...
– Yves Daoust
Dec 3 '18 at 14:30
1
As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
– Yves Daoust
Dec 3 '18 at 14:31
1
This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
– Did
Dec 3 '18 at 14:52
Ask Faulhaber...
– Yves Daoust
Dec 3 '18 at 14:30
Ask Faulhaber...
– Yves Daoust
Dec 3 '18 at 14:30
1
1
As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
– Yves Daoust
Dec 3 '18 at 14:31
As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
– Yves Daoust
Dec 3 '18 at 14:31
1
1
This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
– Did
Dec 3 '18 at 14:52
This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
– Did
Dec 3 '18 at 14:52
add a comment |
1 Answer
1
active
oldest
votes
Hint:
$$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.
Now
$$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
\=4an^3+bn^3-bn^3+cdots$$ and lower order terms. You conclude that
$$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.
answered Dec 3 '18 at 14:36
Yves DaoustYves Daoust
124k671222
add a comment |
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1 Answer
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Hint:
$$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.
Now
$$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
\=4an^3+bn^3-bn^3+cdots$$ and lower order terms. You conclude that
$$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.
answered Dec 3 '18 at 14:36
Yves DaoustYves Daoust
124k671222
add a comment |
Hint:
$$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.
Now
$$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
\=4an^3+bn^3-bn^3+cdots$$ and lower order terms. You conclude that
$$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.
answered Dec 3 '18 at 14:36
Yves DaoustYves Daoust
124k671222
add a comment |
Hint:
$$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.
Now
$$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
\=4an^3+bn^3-bn^3+cdots$$ and lower order terms. You conclude that
$$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.
answered Dec 3 '18 at 14:36
Yves DaoustYves Daoust
124k671222
Hint:
$$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.
Now
$$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
\=4an^3+bn^3-bn^3+cdots$$ and lower order terms. You conclude that
$$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.
answered Dec 3 '18 at 14:36
Yves DaoustYves Daoust
124k671222
answered Dec 3 '18 at 14:36
Yves DaoustYves Daoust
124k671222
answered Dec 3 '18 at 14:36
Yves DaoustYves Daoust
124k671222
answered Dec 3 '18 at 14:36
Yves DaoustYves Daoust
124k671222
124k671222
add a comment |
add a comment |
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Ask Faulhaber...
– Yves Daoust
Dec 3 '18 at 14:30
1
As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
– Yves Daoust
Dec 3 '18 at 14:31
1
This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
– Did
Dec 3 '18 at 14:52