Question

The value of \(\underset{n\rightarrow \infty }{lim}[\frac{1}{1-n^2}+\frac{2}{1-n^2}+\ldots +\frac{n}{1-n^2}]^3\) is

• A. \(8\)
• B. \(-8\)
• C. \(\frac{1}{8}\)
• D. \(\frac{-1}{8}\)

Hint:

Add the numerators with n(n+1)/2, then divide top and bottom by n squared.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
All the fractions share the denominator \(1-n^2\), so the numerators can be added first.

Step 2: Add the numerators
\[ 1+2+\dots+n=\frac{n(n+1)}{2} \]
The bracket becomes
\[ \frac{n(n+1)}{2(1-n^2)} \]

Step 3: Take the limit
Divide the top and bottom by \(n^2\):
\[ \frac{1+\frac1n}{2\left(\frac1{n^2}-1\right)}\to\frac{1}{2(0-1)}=-\frac12 \]

Step 4: Cube
\[ \left(-\frac12\right)^3=-\frac18 \]
This is option (D). The sign matters: because the denominator \(1-n^2\) is negative for large \(n\), the limit is negative, so options (A) and (C) are ruled out.

Final Answer:
The bracket tends to -1/2, so the limit of its cube is -1/8, option (D). \[ \boxed{-\frac{1}{8}} \]