Find `a` for which `lim_(n->oo) (1^a+2^a+3^a+...+n^a)/((n+1)^(a-1)[(na+1)+(na+2)+...+(na+n)])=1/60`

Options

A. 5
B. 7
C. `(-15)/2`
D. `(-17)/2`

Explanation

To solve the problem, we need to find the value of \( a \) such that

\[ \lim_{n \to \infty} \frac{1^a + 2^a + 3^a + \ldots + n^a}{(n+1)^{a-1} \left( (na+1) + (na+2) + \ldots + (na+n) \right)} = \frac{1}{60} \]

### Step 1: Rewrite the numerator
The numerator can be expressed as a summation:

\[ 1^a + 2^a + 3^a + \ldots + n^a = \sum_{r=1}^{n} r^a \]

### Step 2: Simplify the denominator
The denominator can be simplified by factoring out \( n \):

\[ (na + 1) + (na + 2) + \ldots + (na + n) = n \left( a + \frac{1}{n} + \frac{2}{n} + \ldots + 1 \right) = n \left( a + \frac{1}{n} \sum_{k=1}^{n} k \right) \]

Using the formula for the sum of the first \( n \) natural numbers, we have:

\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]

Thus, the denominator becomes:

\[ (n+1)^{a-1} \cdot n \left( a + \frac{1}{n} \cdot \frac{n(n+1)}{2} \right) = (n+1)^{a-1} \cdot n \left( a + \frac{(n+1)}{2} \right) \]

### Step 3: Analyze the limit
Now we can rewrite the limit:

\[ \lim_{n \to \infty} \frac{\sum_{r=1}^{n} r^a}{(n+1)^{a-1} \cdot n \left( a + \frac{(n+1)}{2} \right)} \]

### Step 4: Apply asymptotic behavior
Using the asymptotic behavior of the summation \( \sum_{r=1}^{n} r^a \sim \frac{n^{a+1}}{a+1} \) as \( n \to \infty \):

\[ \lim_{n \to \infty} \frac{\frac{n^{a+1}}{a+1}}{(n+1)^{a-1} \cdot n \left( a + \frac{n}{2} \right)} \]

### Step 5: Simplify further
As \( n \to \infty \):

\[ (n+1)^{a-1} \sim n^{a-1} \]

Thus, we have:

\[ \lim_{n \to \infty} \frac{\frac{n^{a+1}}{a+1}}{n^{a-1} \cdot n \left( a + \frac{n}{2} \right)} = \lim_{n \to \infty} \frac{n^{a+1}}{(a+1) n^{a} \left( a + \frac{n}{2} \right)} \]

This simplifies to:

\[ \lim_{n \to \infty} \frac{1}{(a+1) \left( a + \frac{n}{2} \right)} = \frac{1}{60} \]

### Step 6: Solve for \( a \)
Setting the limit equal to \( \frac{1}{60} \):

\[ \frac{1}{(a+1) \left( a + \frac{n}{2} \right)} = \frac{1}{60} \]

As \( n \to \infty \), we can ignore \( a \) in the denominator:

\[ \frac{1}{(a+1) \cdot \frac{n}{2}} = \frac{1}{60} \]

This gives us:

\[ (a+1) \cdot \frac{n}{2} = 60 \]

### Step 7: Rearranging gives a quadratic equation
Rearranging leads to:

\[ 2(a+1) = 60 \implies a + 1 = 30 \implies a = 29 \]

### Step 8: Verify the solution
To verify, we can substitute \( a = 29 \) back into our limit and check if it holds true.

### Final Answer
Thus, the value of \( a \) is:

\[ \boxed{29} \]