If `A = lim_(n rarr oo) sum_(r = 1)^(n) tan^(-1) ((1)/(2r^(2)))`, then A is equal to

Options

A. `(pi)/(4)`
B. 1
C. `(pi)/(3)`
D. `lt 1`

Answer

To solve the problem, we need to evaluate the limit:

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

### Step 1: Rewrite the term inside the summation
We start by rewriting the term \(\tan^{-1}\left(\frac{1}{2r^2}\right)\) using the identity for the difference of two inverse tangents:

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

We can express \(\tan^{-1}\left(\frac{1}{2r^2}\right)\) in a useful form. We use the identity:

\[ \tan^{-1}(x) = \tan^{-1}(2r + 1) - \tan^{-1}(2r - 1) \]

Thus, we can write:

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

### Step 2: Substitute into the summation
Now, substituting this back into our summation, we get:

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

### Step 3: Recognize the telescoping nature of the series
Notice that this is a telescoping series. When we expand it, we see that most terms will cancel out:

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

### Step 4: Evaluate the limit
As \(n\) approaches infinity, \(\tan^{-1}(2n + 1)\) approaches \(\frac{\pi}{2}\). Therefore, we have:

\[ A = \frac{\pi}{2} - \tan^{-1}(1) \]

Since \(\tan^{-1}(1) = \frac{\pi}{4}\), we can substitute this value in:

\[ A = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Final Result
Thus, the value of \(A\) is:

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