`lim_(n to oo){(1^(m)+2^(m)+3^(m)+...+ n^(m))/(n^(m+1))}` equals

Options

A. `(1)/(m+1)`
B. `(1)/(m+2)`
C. `(1)/(m)`
D. none of these

Explanation

To solve the limit

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

we can follow these steps:

### Step 1: Rewrite the sum
We start by rewriting the sum in the numerator:

\[ S_n = 1^m + 2^m + 3^m + \ldots + n^m. \]

Thus, we can express our limit as:

\[ \lim_{n \to \infty} \frac{S_n}{n^{m+1}}. \]

### Step 2: Factor out \(n^m\)
Notice that we can factor \(n^m\) out of the sum \(S_n\):

\[ S_n = n^m \left(\frac{1^m}{n^m} + \frac{2^m}{n^m} + \frac{3^m}{n^m} + \ldots + \frac{n^m}{n^m}\right) = n^m \left(\frac{1^m}{n^m} + \frac{2^m}{n^m} + \ldots + 1\right). \]

### Step 3: Simplify the limit
Now, substituting this back into our limit gives:

\[ \lim_{n \to \infty} \frac{n^m \left(\frac{1^m}{n^m} + \frac{2^m}{n^m} + \ldots + 1\right)}{n^{m+1}} = \lim_{n \to \infty} \frac{1}{n} \left(\frac{1^m}{n^m} + \frac{2^m}{n^m} + \ldots + 1\right). \]

### Step 4: Recognize the Riemann sum
The expression inside the limit can be recognized as a Riemann sum for the function \(f(x) = x^m\) over the interval \([0, 1]\):

\[ \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}\right)^m. \]

As \(n \to \infty\), this Riemann sum converges to the integral:

\[ \int_0^1 x^m \, dx. \]

### Step 5: Evaluate the integral
Now we compute the integral:

\[ \int_0^1 x^m \, dx = \left[\frac{x^{m+1}}{m+1}\right]_0^1 = \frac{1^{m+1}}{m+1} - \frac{0^{m+1}}{m+1} = \frac{1}{m+1}. \]

### Conclusion
Thus, we find that:

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

So the final answer is:

\[ \frac{1}{m+1}. \]