`lim_(nto oo) overset(n)underset(r=1)Sigma tan^(-1)(1/(2r^2))` is equal to

Options

A. `pi/4`
B. `(pi^2)/4`
C. `pi/2`
D. `(pi^2)/16`

Answer

To solve the limit

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{2r^2}\right), \]

we will follow these steps:

### Step 1: Rewrite the term inside the summation
We start with the term \(\tan^{-1}\left(\frac{1}{2r^2}\right)\). We can rewrite it using the identity for the difference of arctangents:

\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right). \]

We will express \(\tan^{-1}\left(\frac{1}{2r^2}\right)\) in a form that allows us to use this identity.

### Step 2: Use the difference of arctangents
We can manipulate the expression:

\[ \tan^{-1}\left(\frac{1}{2r^2}\right) = \tan^{-1}(2r + 1) - \tan^{-1}(2r - 1). \]

This gives us a telescoping series when we sum over \(r\).

### Step 3: Set up the summation
Now, we can rewrite the limit as:

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

### Step 4: Evaluate the telescoping series
When we expand the summation, we notice that most terms will cancel out:

\[ \left( \tan^{-1}(3) - \tan^{-1}(1) \right) + \left( \tan^{-1}(5) - \tan^{-1}(3) \right) + \left( \tan^{-1}(7) - \tan^{-1}(5) \right) + \ldots + \left( \tan^{-1}(2n + 1) - \tan^{-1}(2n - 1) \right). \]

After cancellation, we are left with:

\[ \tan^{-1}(2n + 1) - \tan^{-1}(1). \]

### Step 5: Take the limit
Now we take the limit as \(n\) approaches infinity:

\[ \lim_{n \to \infty} \left( \tan^{-1}(2n + 1) - \tan^{-1}(1) \right). \]

As \(n \to \infty\), \(\tan^{-1}(2n + 1) \to \frac{\pi}{2}\) and \(\tan^{-1}(1) = \frac{\pi}{4}\). Thus, we have:

\[ \lim_{n \to \infty} \left( \frac{\pi}{2} - \frac{\pi}{4} \right) = \frac{\pi}{4}. \]

### Final Answer
Therefore, the limit is:

\[ \frac{\pi}{4}. \]