`lim_(nto oo)1/n+(1)/(sqrt(n^(2)+n))+(1)/(sqrt(n^(2)+2n))+...(1)/(sqrt(n^(2)+(n-1)n))` is equal to

Options

A. `2+2sqrt(2)`
B. `2sqrt(2)-2`
C. `2sqrt(2)`
D. 2

Explanation

To solve the limit

\[ \lim_{n \to \infty} \left( \frac{1}{n} + \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+2n}} + \ldots + \frac{1}{\sqrt{n^2+(n-1)n}} \right), \]

we can break it down step by step.

### Step 1: Rewrite the Expression
We start by rewriting the limit expression in a more manageable form. Notice that each term in the sum can be expressed as:

\[ \frac{1}{\sqrt{n^2 + kn}} = \frac{1}{n\sqrt{1 + \frac{k}{n}}} \quad \text{for } k = 1, 2, \ldots, n-1. \]

Thus, we can rewrite the limit as:

\[ \lim_{n \to \infty} \left( \frac{1}{n} + \sum_{k=1}^{n-1} \frac{1}{n\sqrt{1 + \frac{k}{n}}} \right). \]

### Step 2: Factor Out \(\frac{1}{n}\)
Now, factor out \(\frac{1}{n}\):

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + \sum_{k=1}^{n-1} \frac{1}{\sqrt{1 + \frac{k}{n}}} \right). \]

### Step 3: Convert the Sum to an Integral
As \(n\) approaches infinity, the sum can be approximated by an integral. The term \(\frac{1}{n}\) can be interpreted as \(dx\) where \(x = \frac{k}{n}\). Hence, the sum becomes:

\[ \sum_{k=1}^{n-1} \frac{1}{\sqrt{1 + \frac{k}{n}}} \approx n \int_0^1 \frac{1}{\sqrt{1+x}} \, dx. \]

### Step 4: Evaluate the Integral
Now we need to evaluate the integral:

\[ \int_0^1 \frac{1}{\sqrt{1+x}} \, dx. \]

Using the substitution \(u = 1 + x\), we have \(du = dx\) and the limits change from \(x=0\) to \(x=1\) which corresponds to \(u=1\) to \(u=2\):

\[ \int_1^2 \frac{1}{\sqrt{u}} \, du = [2\sqrt{u}]_1^2 = 2\sqrt{2} - 2. \]

### Step 5: Combine Results
Now substituting back into our limit expression:

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + n \cdot (2\sqrt{2} - 2) \right) = \lim_{n \to \infty} \left( \frac{1}{n} + 2\sqrt{2} - 2 \right) = 2\sqrt{2} - 2. \]

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

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