`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 of the given summation, we will follow a systematic approach. Let's break it down step by step.

### Step 1: Write the limit and summation
We start with the expression:
\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^{\frac{1}{a}} \left( n^{a - \frac{1}{a}} + k^{a - \frac{1}{a}} \right)}{n^{a + 1}} \]

### Step 2: Simplify the expression
We can separate the terms in the numerator:
\[ = \lim_{n \to \infty} \sum_{k=1}^{n} \left( \frac{k^{\frac{1}{a}} n^{a - \frac{1}{a}}}{n^{a + 1}} + \frac{k^{\frac{1}{a}} k^{a - \frac{1}{a}}}{n^{a + 1}} \right) \]
This simplifies to:
\[ = \lim_{n \to \infty} \sum_{k=1}^{n} \left( \frac{k^{\frac{1}{a}}}{n^{1 + \frac{1}{a}}} + \frac{k^{a}}{n^{a + 1}} \right) \]

### Step 3: Factor out constants
We can factor out the constants from the summation:
\[ = \lim_{n \to \infty} \left( \frac{1}{n^{1 + \frac{1}{a}}} \sum_{k=1}^{n} k^{\frac{1}{a}} + \frac{1}{n^{a + 1}} \sum_{k=1}^{n} k^{a} \right) \]

### Step 4: Evaluate the sums
Using the formula for the sum of powers:
\[ \sum_{k=1}^{n} k^p \approx \frac{n^{p+1}}{p+1} \quad \text{as } n \to \infty \]
we can write:
\[ \sum_{k=1}^{n} k^{\frac{1}{a}} \approx \frac{n^{1 + \frac{1}{a}}}{1 + \frac{1}{a}} \quad \text{and} \quad \sum_{k=1}^{n} k^{a} \approx \frac{n^{a + 1}}{a + 1} \]

### Step 5: Substitute back into the limit
Substituting these approximations back into the limit gives:
\[ = \lim_{n \to \infty} \left( \frac{1}{n^{1 + \frac{1}{a}}} \cdot \frac{n^{1 + \frac{1}{a}}}{1 + \frac{1}{a}} + \frac{1}{n^{a + 1}} \cdot \frac{n^{a + 1}}{a + 1} \right) \]
This simplifies to:
\[ = \lim_{n \to \infty} \left( \frac{1}{1 + \frac{1}{a}} + \frac{1}{a + 1} \right) \]

### Step 6: Combine the results
Now we can combine the results:
\[ = \frac{1}{1 + \frac{1}{a}} + \frac{1}{a + 1} = \frac{a}{a + 1} + \frac{1}{a + 1} = \frac{a + 1}{a + 1} = 1 \]

### Final Result
Thus, we find that:
\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^{\frac{1}{a}} \left( n^{a - \frac{1}{a}} + k^{a - \frac{1}{a}} \right)}{n^{a + 1}} = 1 \]