`lim_(n->oo)(1^(99)+2^(99)+3^(99)+.......n^(99))/(n^(100))=`

Options

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

Answer

To solve the limit

\[ \lim_{n \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \ldots + n^{99}}{n^{100}}, \]

we will follow these steps:

### Step 1: Recognize the Summation
The numerator consists of the sum of the first \( n \) integers raised to the 99th power. We can denote this summation as:

\[ S_n = 1^{99} + 2^{99} + 3^{99} + \ldots + n^{99}. \]

### Step 2: Use the Formula for the Sum of Powers
The sum of the first \( n \) integers raised to the \( k \)-th power can be approximated using the formula:

\[ \sum_{k=1}^{n} k^p \sim \frac{n^{p+1}}{p+1} \quad \text{as } n \to \infty. \]

For our case, \( p = 99 \):

\[ S_n \sim \frac{n^{100}}{100} \quad \text{as } n \to \infty. \]

### Step 3: Substitute Back into the Limit
Now we can substitute this approximation into our limit:

\[ \lim_{n \to \infty} \frac{S_n}{n^{100}} \sim \lim_{n \to \infty} \frac{\frac{n^{100}}{100}}{n^{100}}. \]

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

\[ \lim_{n \to \infty} \frac{1}{100} = \frac{1}{100}. \]

### Step 5: Conclusion
Thus, the final result is:

\[ \lim_{n \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \ldots + n^{99}}{n^{100}} = \frac{1}{100}. \]

### Final Answer
The answer is

\[ \frac{1}{100}. \]

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