`underset(n to oo)lim {1/(1-n^(2))+(2)/(1-n^(2))+....+(n)/(1-n^(2))}` is equal to

Options

A. 0
B. `-1/2`
C. `1/2`
D. None of these

Answer

To solve the limit

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

we can follow these steps:

### Step 1: Factor out the common denominator
The expression can be rewritten by factoring out \( \frac{1}{1 - n^2} \):

\[ \lim_{n \to \infty} \frac{1}{1 - n^2} \left( 1 + 2 + 3 + \cdots + n \right). \]

### Step 2: Use the formula for the sum of the first \( n \) natural numbers
The sum \( 1 + 2 + 3 + \cdots + n \) is given by the formula:

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

Substituting this into our limit gives:

\[ \lim_{n \to \infty} \frac{1}{1 - n^2} \cdot \frac{n(n + 1)}{2}. \]

### Step 3: Simplify the expression
Now we can simplify the expression:

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

### Step 4: Rewrite the denominator
The denominator \( 1 - n^2 \) can be factored as \( -(n^2 - 1) \):

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

### Step 5: Factor out \( n^2 \) from the denominator
We can factor \( n^2 \) out of the denominator:

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

### Step 6: Simplify further
Now, simplifying the limit gives:

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

### Step 7: Evaluate the limit
As \( n \to \infty \), \( \frac{1}{1 - \frac{1}{n^2}} \) approaches 1, and \( \frac{(n + 1)}{-2n} \) approaches \( \frac{1}{-2} \):

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

Thus, the limit evaluates to:

\[ -\frac{1}{2}. \]

### Final Answer
Therefore, the final answer is:

\[ \boxed{-\frac{1}{2}}. \]

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