`lim_(n->oo)1/nsum_(r=1)^(2n)r/(sqrt(n^2+r^2))` equals

Answer

To solve the limit \( \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{2n} \frac{r}{\sqrt{n^2 + r^2}} \), we can follow these steps:

### Step 1: Rewrite the Summation as an Integral
We start by recognizing that as \( n \) approaches infinity, the sum can be approximated by an integral. We rewrite the sum:

\[ \frac{1}{n} \sum_{r=1}^{2n} \frac{r}{\sqrt{n^2 + r^2}} \approx \int_{0}^{2} \frac{x}{\sqrt{1 + x^2}} \, dx \]

where we have made the substitution \( x = \frac{r}{n} \), which implies \( r = nx \) and \( dr = n \, dx \).

### Step 2: Change the Variable
Next, we change the variable in the integral. We factor out \( n^2 \) from the square root in the denominator:

\[ \sqrt{n^2 + r^2} = n \sqrt{1 + \left(\frac{r}{n}\right)^2} = n \sqrt{1 + x^2} \]

Thus, we can rewrite the expression inside the sum:

\[ \frac{r}{\sqrt{n^2 + r^2}} = \frac{nx}{n\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}} \]

### Step 3: Set Up the Integral
Now, we can express the limit as an integral:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{2n} \frac{r}{\sqrt{n^2 + r^2}} \approx \int_{0}^{2} \frac{x}{\sqrt{1 + x^2}} \, dx \]

### Step 4: Evaluate the Integral
To evaluate the integral, we can use the substitution \( t = 1 + x^2 \), which gives \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \). The limits change as follows:
- When \( x = 0 \), \( t = 1 \)
- When \( x = 2 \), \( t = 5 \)

Thus, the integral becomes:

\[ \int_{1}^{5} \frac{1}{\sqrt{t}} \cdot \frac{dt}{2} = \frac{1}{2} \int_{1}^{5} t^{-1/2} \, dt \]

### Step 5: Solve the Integral
Now we compute the integral:

\[ \frac{1}{2} \left[ 2t^{1/2} \right]_{1}^{5} = \left[ t^{1/2} \right]_{1}^{5} = \sqrt{5} - \sqrt{1} = \sqrt{5} - 1 \]

### Final Result
Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{2n} \frac{r}{\sqrt{n^2 + r^2}} = \sqrt{5} - 1 \]