`lim_(n to oo)[(n+1)/(n^(2)+1^(2))+(n+2)/(n^(2)+2^(2))+....+(1)/(n)]`

Options

A. `(1)/(2)log 2`
B. `-(pi)/(4)+(1)/(2)log 2`
C. `pi + (1)/(2) log 2`
D. `(pi)/(4)+(1)/(2) log 2`

Answer

To solve the limit

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

we can rewrite this expression in summation form:

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

### Step 1: Rewrite the terms

We can factor \(n^2\) out of the denominator:

\[ \frac{n+r}{n^2 + r^2} = \frac{n+r}{n^2(1 + \frac{r^2}{n^2})} = \frac{1 + \frac{r}{n}}{n(1 + \frac{r^2}{n^2})}. \]

### Step 2: Express the limit as a Riemann sum

Now we can express the sum as:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1 + \frac{r}{n}}{n(1 + \frac{r^2}{n^2})}. \]

As \(n\) approaches infinity, \(\frac{r}{n}\) approaches \(x\) where \(x\) ranges from \(0\) to \(1\) as \(r\) goes from \(1\) to \(n\). Therefore, we can express this sum as a Riemann sum:

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

### Step 3: Change of variables

Let \(x = \frac{r}{n}\). Then \(dx = \frac{1}{n}\) and the limits of integration change from \(0\) to \(1\):

\[ \int_{0}^{1} \frac{1+x}{1+x^2} \, dx. \]

### Step 4: Split the integral

We can split the integral:

\[ \int_{0}^{1} \frac{1+x}{1+x^2} \, dx = \int_{0}^{1} \frac{1}{1+x^2} \, dx + \int_{0}^{1} \frac{x}{1+x^2} \, dx. \]

### Step 5: Evaluate the first integral

The first integral can be evaluated as:

\[ \int_{0}^{1} \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \bigg|_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Step 6: Evaluate the second integral

For the second integral, we use the substitution \(u = 1 + x^2\), which gives \(du = 2x \, dx\) or \(dx = \frac{du}{2x}\):

\[ \int_{0}^{1} \frac{x}{1+x^2} \, dx = \frac{1}{2} \int_{1}^{2} \frac{1}{u} \, du = \frac{1}{2} \ln(u) \bigg|_{1}^{2} = \frac{1}{2} (\ln(2) - \ln(1)) = \frac{1}{2} \ln(2). \]

### Step 7: Combine the results

Combining both results, we have:

\[ \int_{0}^{1} \frac{1+x}{1+x^2} \, dx = \frac{\pi}{4} + \frac{1}{2} \ln(2). \]

### Final Answer

Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \left[ \frac{n+1}{n^2 + 1^2} + \frac{n+2}{n^2 + 2^2} + \ldots + \frac{1}{n} \right] = \frac{\pi}{4} + \frac{1}{2} \ln(2). \]