Using the definite integral, find the limit of the sum

Using the definite integral, find the limit of the sum: $\lim_{n \rightarrow \infty}\left(\frac{1^p}{n^{p+1}}+\frac{2^p}{n^{p+1}}+\cdots+\frac{n^p}{n^{p+1}}\right)$, for $p>0$.

We will express this sum as a Riemann sum and find its limit using the definite integral.

Step 1: Rewrite the sum

The sum can be written as:

\[S_n = \sum_{k=1}^{n} \frac{k^p}{n^{p+1}} = \frac{1}{n^{p+1}} \sum_{k=1}^{n} k^p\]

Alternatively, we can factor:

\[S_n = \sum_{k=1}^{n} \frac{k^p}{n^{p+1}} = \sum_{k=1}^{n} \frac{k^p}{n^p \cdot n} = \sum_{k=1}^{n} \left(\frac{k}{n}\right)^p \cdot \frac{1}{n}\]

Step 2: Recognize as a Riemann sum

Let $\Delta x = \frac{1}{n}$ and $x_k = \frac{k}{n}$ where $k = 1, 2, \ldots, n$.

Then:

\[S_n = \sum_{k=1}^{n} \left(\frac{k}{n}\right)^p \cdot \frac{1}{n} = \sum_{k=1}^{n} (x_k)^p \cdot \Delta x\]

This is a Riemann sum for the function $f(x) = x^p$ over an interval.

Step 3: Determine the interval

The terms correspond to: $x_1 = \frac{1}{n}$ (first term) and $x_n = \frac{n}{n} = 1$ (last term).

As $n \to \infty$, $x_1 = \frac{1}{n} \to 0$ and $x_n = 1$ remains 1.

Therefore, this is a Riemann sum for $f(x) = x^p$ over the interval $[0, 1]$.

Step 4: Take the limit

As $n \to \infty$, the Riemann sum converges to the definite integral:

\[\lim_{n \to \infty} S_n = \int_0^1 x^p \, dx\]

Step 5: Evaluate the integral

For $p > 0$ (so $p \neq -1$), the antiderivative of $x^p$ is $\frac{x^{p+1}}{p+1}$:

\[\int_0^1 x^p \, dx = \left[\frac{x^{p+1}}{p+1}\right]_0^1\]

\[= \frac{1^{p+1}}{p+1} - \frac{0^{p+1}}{p+1}\]

\[= \frac{1}{p+1} - 0\]

\[= \frac{1}{p+1}\]

Verification for specific values:

• Case $p = 1$: \[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2} = \int_0^1 x \, dx = \frac{1}{2} \] Using the formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$: \[ \frac{1}{n^2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{2n^2} = \frac{n+1}{2n} \to \frac{1}{2} \]
• Case $p = 2$: \[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^2}{n^3} = \int_0^1 x^2 \, dx = \frac{1}{3} \] Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$: \[ \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{6n^3} \to \frac{2n^3}{6n^3} = \frac{1}{3} \]

Geometric interpretation:

The sum $\sum_{k=1}^{n} \frac{k^p}{n^{p+1}}$ represents a Riemann sum approximation of the area under the curve $y = x^p$ from $x = 0$ to $x = 1$.

For different values of $p$: when $p=1$, the curve is $y=x$ (a straight line) with area $\frac12$; when $01$, the curve is concave up so the area is less than $\frac12$. As $p$ increases, the curve $y=x^p$ stays closer to the $x$-axis in $(0,1)$, so the area $\frac{1}{p+1}$ decreases.

Final answer:

\[\boxed{\lim_{n \rightarrow \infty}\left(\frac{1^p}{n^{p+1}}+\frac{2^p}{n^{p+1}}+\cdots+\frac{n^p}{n^{p+1}}\right) = \frac{1}{p+1} \quad (p > 0)}\]