lim (n→∞) (1/(1-n²) + 2/(1-n²) + ... + n/(1-n²)) is equal to (a) 0 (b) -1/2 (c) 1/2 (d) none of thes

Q: Wait, can't I just take the limit of each term individually and add them up?

That is a classic trap! The limit of a sum is the sum of the limits only if the number of terms is fixed. Here, the number of terms $n$ is growing to infinity, so you must simplify the expression first.

Q: Why does $1-n^2$ become $(1-n)(1+n)$? Isn't it $(n-1)(n+1)$?

Both are correct! Since $(1-n) = -(n-1)$, you just need to be careful with the signs. Using $(1-n)(1+n)$ is perfectly valid and helps cancel the $(n+1)$ term directly.

Q: Why do we divide by $n$ at the end? Is there another way?

Dividing by the highest power of $n$ is the most robust method for rational functions at infinity. You could also use L'Hôpital's Rule, but algebraic simplification is often faster and less prone to calculation errors.

Question

$\lim_{n 	o \infty} \left( \frac{1}{1-n^2} + \frac{2}{1-n^2} + \dots + \frac{n}{1-n^2} ight)$ is equal to

Options:

(A) 0

(B) $-1/2$

(C) $1/2$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The beauty of mathematics lies in its ability to find order within chaos. Let us examine the expression:

$$\lim_{n 	o \infty} \left( \frac{1}{1-n^2} + \frac{2}{1-n^2} + \dots + \frac{n}{1-n^2} ight)$$

The first thing that should strike you is that the denominator is identical for every single term. This is a gift! We do not need to hunt for a common denominator; it is already staring us in the face.

By combining the numerators, we transform our expression into:

$$\lim_{n 	o \infty} \frac{1 + 2 + 3 + \dots + n}{1-n^2}$$

### The Power of Summation

Now, look at the numerator. It is the sum of the first $n$ natural numbers, which is a fundamental building block of series. We know that:

$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$

Substituting this into our expression, we get:

$$\lim_{n 	o \infty} \frac{n(n+1)}{2(1-n^2)}$$

This formula is a powerful tool that collapses a long, intimidating series into a compact algebraic term. It acts as the bridge between the discrete and the continuous.

### The Algebraic Dance

Now, we enter the realm of algebraic manipulation. The denominator $1-n^2$ is a classic difference of squares, which we can factor as $(1-n)(1+n)$.

Our expression becomes:

$$\lim_{n 	o \infty} \frac{n(n+1)}{2(1-n)(1+n)}$$

Notice the magic? The term $(n+1)$ appears in both the numerator and the denominator. We can cancel them out to obtain:

$$\lim_{n 	o \infty} \frac{n}{2(1-n)}$$

### The Infinity Perspective

To evaluate the limit as $n 	o \infty$, we use the standard JEE technique for rational functions at infinity: divide the numerator and denominator by the highest power of $n$, which is $n^1$.

This gives us:

$$\lim_{n 	o \infty} \frac{1}{2(\frac{1}{n} - 1)}$$

As $n$ approaches infinity, the term $\frac{1}{n}$ vanishes to zero. We are left with:

$$\frac{1}{2(0-1)} = -\frac{1}{2}$$

This journey shows that even complex-looking problems are just a series of simple, elegant steps. The final answer is **$-\frac{1}{2}$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{n \to \infty} (\frac{1}{1 - n ^{2}} + \frac{2}{1 - n ^{2}} + \dots + \frac{n}{1 - n ^{2}})$ is equal to

Select Answer:

Visualized Solution

Identifying the Common Denominator

Sum of First $n$ Natural Numbers

Substituting the Sum Formula

Factoring the Denominator

Canceling Common Terms

Preparing for Limit at Infinity

Evaluating the Limit

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Power of Summation

The Algebraic Dance

The Infinity Perspective

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