`lim_(n rarr oo) (1^(99) + 2^(99) + …+ n^(99))/(n^(100))=`

Options

A. `(99)/(100)`
B. `(1)/(100)`
C. `(1)/(99)`
D. `(1)/(101)`

Answer

To solve the limit

\[ \lim_{n \to \infty} \frac{1^{99} + 2^{99} + \ldots + n^{99}}{n^{100}}, \]

we can follow these steps:

### Step 1: Identify the Sum
We need to analyze the numerator, which is the sum of the first \( n \) integers raised to the power of 99. This can be expressed as:

\[ S_n = 1^{99} + 2^{99} + \ldots + n^{99}. \]

### Step 2: Use the Formula for the Sum of Powers
There is a known formula for the sum of the first \( n \) integers raised to the \( k \)-th power:

\[ S_n = \frac{n^{k+1}}{k+1} + O(n^k), \]

where \( O(n^k) \) represents lower order terms. For \( k = 99 \), we have:

\[ S_n = \frac{n^{100}}{100} + O(n^{99}). \]

### Step 3: Substitute into the Limit
Now we substitute \( S_n \) back into our limit:

\[ \lim_{n \to \infty} \frac{S_n}{n^{100}} = \lim_{n \to \infty} \frac{\frac{n^{100}}{100} + O(n^{99})}{n^{100}}. \]

### Step 4: Simplify the Expression
This simplifies to:

\[ \lim_{n \to \infty} \left( \frac{1}{100} + \frac{O(n^{99})}{n^{100}} \right). \]

### Step 5: Analyze the Limit
As \( n \to \infty \), the term \( \frac{O(n^{99})}{n^{100}} \) approaches 0, since the degree of the numerator is less than the degree of the denominator. Thus, we have:

\[ \lim_{n \to \infty} \left( \frac{1}{100} + 0 \right) = \frac{1}{100}. \]

### Final Result
Therefore, the limit is:

\[ \lim_{n \to \infty} \frac{1^{99} + 2^{99} + \ldots + n^{99}}{n^{100}} = \frac{1}{100}. \]

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