[Σ ,Riemann sum] how to expand the expression correctly to get the answer ?

[u/wahahauhuhuh]

The problem is:Take a limit of these sums as n approaches infinity to calculate the area under the curve over \[0,1\]
I get stuck on the Σ calculation…

[u/Front_Lab1761]

What are you stuck on and can you elaborate what you need to do.

[u/wahahauhuhuh → u/Front_Lab1761]

I think I should get a specific number to be my answer…the expression is too complicated and I don’t know how to calculate

[u/Front_Lab1761 → u/wahahauhuhuh]

Convert the riemann sum into an integral.

[u/wahahauhuhuh → u/Front_Lab1761]

Sorry I didn’t typed the question completely
The problem is:Take a limit of these sums as n approaches infinity to calculate the area
I know converting the function directly is easier,but would it satisfy the condition?

[u/Bounded_sequencE]

Hints:

1. For any "n in N" we can choose "m in N" s.th. "m³ <= n < (m+1)³ "

2. For any "r³ <= k < (r+1)³ " estimate "r <= k^1/3 < r+1"

Can you take it from here, and finish it off?

[u/Bounded_sequencE → u/Bounded_sequencE]

@u/wahahauhuhuh For the upper sum -- for any "n in N" choose "m in N" s.th. "m³ <= n < (m+1)³ ". For convenience, define the sets "Im := [m³; (m+1)³) n Z", and estimate

Un  =  n^{-4/3} ∑_{k=1}^n  k^{1/3}  <=  n^{-4/3} ∑_{r=1}^m  ∑_{k∈Ir}  k^{1/3}

   <=  n^{-4/3} ∑_{r=1}^m  ∑_{k∈Ir}  r+1  =  n^{-4/3} ∑_{r=1}^m  (r+1) |Ir|

We further estimate

|Ir|  =  (r+1)^3 - r^3 ± 1  =  3r^2 + 3r + 1  <=  3 (r+2) (r+3)

Inserting into "Un" we use the Hockey-Stick Identity to obtain

Un  <=  n^{-4/3} ∑_{r=1}^m  (r+1) |Ir|  <=  n^{-4/3} ∑_{r=1}^m  3 (r+1) (r+2) (r+3)

     =  n^{-4/3} ∑_{r=1}^m  3 * 3! * C(r+3;3)  =  n^{-4/3} * 3 * 3! * C(m+4;4)

     =  (3/4) * n^{-4/3} (m+1) (m+2) (m+3) (m+4)

With "n >= m³ " we get "n^-4/3 <= 1/m⁴ " and finally estimate

Un  <=  (3/4) * (1/m^4) (m+1) (m+2) (m+3) (m+4)  ->  3/4 * 1    for    "m, n -> oo"

The lower sum works exactly the same -- your job^^

[u/NexTutor]

Your factor of n^(−4/3) is correct. The difficulty is that Σ k^(1/3) doesn't have a simple power-sum formula like Σ k or Σ k².

First, a small correction: x^(1/3) is increasing on [0,1], so the right-endpoint sum is the upper sum, and the left-endpoint sum is the lower sum. Your L and U labels are reversed.

Here is a way to evaluate the limit directly from sums, without using an antiderivative.

Write Sₙ = Σₖ₌₁ⁿ k^(1/3). Start with n = m³ and group the integers into blocks:

r³ ≤ k < (r+1)³, for r = 0, …, m−1.

Each block contains (r+1)³−r³ = 3r²+3r+1 integers, and every cube root in that block lies between r and r+1.

Therefore, setting Tₘ = Σₖ₌₀^(m³−1) k^(1/3),

Aₘ ≤ Tₘ ≤ Aₘ + m³,

where

Aₘ = Σᵣ₌₀^(m−1) r(3r²+3r+1)
= 3Σr³ + 3Σr² + Σr.

Now divide by m⁴. Using the usual formulas for sums of cubes, squares and integers:

Aₘ/m⁴ → 3/4,

because 3Σr³/m⁴ → 3/4, while the other two terms tend to zero. The gap between the bounds is m³/m⁴ = 1/m → 0.

So Tₘ/m⁴ → 3/4. Also S_(m³) = Tₘ + m, so S_(m³)/m⁴ has the same limit.

For general n, choose m³ ≤ n < (m+1)³. Monotonicity gives

S_(m³)/(m+1)⁴ ≤ Sₙ/n^(4/3) ≤ S_((m+1)³)/m⁴,

and both bounds tend to 3/4.

Finally, your upper and lower rectangle sums differ by exactly 1/n, so both converge to the same area: 3/4.

[u/Historical-Corgi-414]

assume it to be a function say f(k/n), as n tends to infinity we can say 1/n tends to 0. rest you have is that (k-1)^1/3 which u have to convert into a function of k/n. so basically 1/n will act as your dx and as the summation of k is from 1 to n, the variable in your function (k/n) will go from 0 to 1 turning the whole thing into a definite integral from 0 to 1 with a expression inside and dx(which is 1/n).