Value of `lim_(n to oo) ((n+1)^(1//3)+(n+2)^(1//3)+…+(2n)^(1//3))/(n^(4//3))` is equal to

Options

A. `(1)/(4)(2^(1//4)-1)`
B. `(3)/(4)(2^(1//4)-1)`
C. `(3)/(4)(2^(4//3)-1)`
D. `(3)/(4)(2^(4//3)+1)`

Answer

To solve the limit problem

\[ \lim_{n \to \infty} \frac{(n+1)^{1/3} + (n+2)^{1/3} + \ldots + (2n)^{1/3}}{n^{4/3}}, \]

we can follow these steps:

### Step 1: Rewrite the sum
The expression in the numerator can be rewritten as a sum:

\[ \sum_{k=1}^{n} (n+k)^{1/3}. \]

### Step 2: Factor out \(n^{1/3}\)
We can factor \(n^{1/3}\) out of each term in the sum:

\[ \sum_{k=1}^{n} (n+k)^{1/3} = \sum_{k=1}^{n} n^{1/3} \left(1 + \frac{k}{n}\right)^{1/3} = n^{1/3} \sum_{k=1}^{n} \left(1 + \frac{k}{n}\right)^{1/3}. \]

### Step 3: Substitute into the limit
Substituting this back into our limit gives:

\[ \lim_{n \to \infty} \frac{n^{1/3} \sum_{k=1}^{n} \left(1 + \frac{k}{n}\right)^{1/3}}{n^{4/3}} = \lim_{n \to \infty} \frac{\sum_{k=1}^{n} \left(1 + \frac{k}{n}\right)^{1/3}}{n}. \]

### Step 4: Recognize the Riemann Sum
The expression \(\frac{1}{n} \sum_{k=1}^{n} \left(1 + \frac{k}{n}\right)^{1/3}\) is a Riemann sum for the function \(f(x) = (1+x)^{1/3}\) over the interval \([0, 1]\). Therefore, as \(n \to \infty\), this sum approaches the integral:

\[ \int_0^1 (1+x)^{1/3} \, dx. \]

### Step 5: Calculate the integral
Now we compute the integral:

\[ \int_0^1 (1+x)^{1/3} \, dx. \]

Using the substitution \(u = 1+x\), \(du = dx\), the limits change from \(x=0\) to \(x=1\) into \(u=1\) to \(u=2\):

\[ \int_1^2 u^{1/3} \, du. \]

Calculating this integral gives:

\[ \left[ \frac{u^{4/3}}{4/3} \right]_1^2 = \frac{3}{4} \left(2^{4/3} - 1^{4/3}\right) = \frac{3}{4} (2^{4/3} - 1). \]

### Step 6: Final result
Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \frac{(n+1)^{1/3} + (n+2)^{1/3} + \ldots + (2n)^{1/3}}{n^{4/3}} = \frac{3}{4} (2^{4/3} - 1). \]