Using the definite integral, find the limit of the sum

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

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:

\[S_n = \sum_{k=n}^{2n} \frac{1}{k} = \sum_{k=0}^{n} \frac{1}{n+k}\]

We can factor out $\frac{1}{n}$:

\[S_n = \sum_{k=0}^{n} \frac{1}{n+k} = \sum_{k=0}^{n} \frac{1}{n\left(1+\frac{k}{n}\right)} = \sum_{k=0}^{n} \frac{1}{n} \cdot \frac{1}{1+\frac{k}{n}}\]

Step 2: Recognize as a Riemann sum

Let $\Delta x = \frac{1}{n}$ and $x_k = \frac{k}{n}$ where $k = 0, 1, 2, \ldots, n$.

Then:

\[S_n = \sum_{k=0}^{n} \frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n} = \sum_{k=0}^{n} \frac{1}{1+x_k} \cdot \Delta x\]

This is a Riemann sum for the function $f(x) = \frac{1}{1+x}$ over the interval $[0, 1]$.

Step 3: Take the limit

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

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

Step 4: Evaluate the integral

We know that:

\[\int \frac{1}{1+x} \, dx = \ln|1+x| + C\]

Applying the Newton-Leibniz formula:

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

\[= \ln(1+1) - \ln(1+0)\]

\[= \ln(2) - \ln(1)\]

\[= \ln(2) - 0\]

\[= \ln(2)\]

Verification of the setup:

Let's verify that our Riemann sum is correct by checking the endpoints: when $k=0$, $\frac{1}{n+0}=\frac1n$; when $k=n$, $\frac{1}{n+n}=\frac1{2n}$.

The sum has $n+1$ terms (from $k=0$ to $k=n$), which corresponds to the right Riemann sum with $n+1$ subintervals. As $n \to \infty$, this converges to the same limit.

Geometric interpretation:

The sum $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}$ represents the sum of reciprocals of consecutive integers from $n$ to $2n$. As $n$ increases, each term becomes smaller but we're adding more terms; the Riemann sum interpretation shows that the area under the curve $y = \frac{1}{1+x}$ from $x=0$ to $x=1$ captures this balance. The result $\ln(2) \approx 0.693$ shows that the sum approaches a finite limit.

Connection to harmonic series:

This result is related to the harmonic series $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$.

Our sum can be written as:

\[S_n = H_{2n} - H_{n-1} = (H_{2n} - H_n) + (H_n - H_{n-1})\]

As $n \to \infty$, the term $(H_n - H_{n-1}) = \frac{1}{n} \to 0$, and $(H_{2n} - H_n) \to \ln(2)$.

Final answer:

\[\boxed{\lim_{n \rightarrow \infty}\left(\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\right) = \ln(2)}\]