`lim_(nrarroo) sum_(k=1)^(n)(k^(1//a{n^(a-(1)/(a))+k^(a-(1)/(a))}))/(n^(a+1))` is equal to

Options

A. 1
B. 2
C. 43467
D. 4

Explanation

To solve the limit problem given, we will follow a structured approach. The problem is:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^{\frac{1}{a}} \left(n^{a - \frac{1}{a}} + k^{a - \frac{1}{a}}\right)}{n^{a + 1}} \]

### Step 1: Rewrite the expression
We can rewrite the expression inside the limit for clarity:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^{\frac{1}{a}} n^{a - \frac{1}{a}} + k^{\frac{1}{a}} k^{a - \frac{1}{a}}}{n^{a + 1}} \]

### Step 2: Factor out common terms
Notice that we can factor out \( \frac{1}{n^{a + 1}} \):

\[ = \lim_{n \to \infty} \frac{1}{n^{a + 1}} \sum_{k=1}^{n} \left( k^{\frac{1}{a}} n^{a - \frac{1}{a}} + k^{a} \right) \]

### Step 3: Split the sum
We can split the summation into two parts:

\[ = \lim_{n \to \infty} \frac{1}{n^{a + 1}} \left( n^{a - \frac{1}{a}} \sum_{k=1}^{n} k^{\frac{1}{a}} + \sum_{k=1}^{n} k^{a} \right) \]

### Step 4: Analyze the first sum
The first sum can be approximated using the formula for the sum of powers:

\[ \sum_{k=1}^{n} k^{\frac{1}{a}} \approx \frac{n^{\frac{1}{a} + 1}}{\frac{1}{a} + 1} \]

### Step 5: Analyze the second sum
Similarly, for the second sum:

\[ \sum_{k=1}^{n} k^{a} \approx \frac{n^{a + 1}}{a + 1} \]

### Step 6: Substitute back into the limit
Substituting these approximations back into our limit gives:

\[ = \lim_{n \to \infty} \frac{1}{n^{a + 1}} \left( n^{a - \frac{1}{a}} \cdot \frac{n^{\frac{1}{a} + 1}}{\frac{1}{a} + 1} + \frac{n^{a + 1}}{a + 1} \right) \]

### Step 7: Simplify the expression
Now we simplify each term:

1. For the first term:
\[ n^{a - \frac{1}{a}} \cdot \frac{n^{\frac{1}{a} + 1}}{\frac{1}{a} + 1} = \frac{n^{a + 1}}{\frac{1}{a} + 1} \]

2. For the second term:
\[ \frac{n^{a + 1}}{a + 1} \]

Combining these gives:

\[ = \lim_{n \to \infty} \frac{n^{a + 1}}{n^{a + 1}} \left( \frac{1}{\frac{1}{a} + 1} + \frac{1}{a + 1} \right) \]

### Step 8: Evaluate the limit
As \( n \to \infty \), the \( n^{a + 1} \) terms cancel out:

\[ = \frac{1}{\frac{1}{a} + 1} + \frac{1}{a + 1} \]

### Step 9: Combine the fractions
Finding a common denominator:

\[ = \frac{(a + 1) + (1)}{(a + 1)(\frac{1}{a} + 1)} = \frac{a + 2}{(a + 1)(\frac{1 + a}{a})} = \frac{a(a + 2)}{(a + 1)(a + 1)} = \frac{a(a + 2)}{(a + 1)^2} \]

### Final Result
Thus, the limit evaluates to:

\[ \frac{a(a + 2)}{(a + 1)^2} \]