Problem 29 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty}\left[\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots+\frac{1}{(2 n+1)(2 n+3)}\right]\) is equal to (A) 1 (B) \(\frac{1}{2}\) (C) \(-\frac{1}{2}\) (D) None of these

Step-by-Step Solution

Answer

The limit is \(\frac{1}{2}\). (Option B)

Step 1: Understand the Series

The given expression is a series where each term is of the form \(\frac{1}{(2n-1)(2n+1)}\). Notice it can be rewritten using partial fractions.

Step 2: Apply Partial Fraction Decomposition

We rewrite each term \(\frac{1}{(2k-1)(2k+1)}\) using partial fractions as \(A/(2k-1) + B/(2k+1)\). Solving for \(A\) and \(B\), we find \(A = 1/2\) and \(B = -1/2\). Thus, \(\frac{1}{(2k-1)(2k+1)} = \frac{1}{2(2k-1)} - \frac{1}{2(2k+1)}\).

Step 3: Recognize the Telescoping Series

The series can now be seen as telescoping: each fraction in the form \(\frac{1}{2(2k-1)} - \frac{1}{2(2k+1)}\) will cancel sequentially with parts of adjacent terms.

Step 4: Identify Surviving Terms

When writing out the terms and cancelling the middle terms, the first term's \(\frac{1}{2*1}\) and the last term's \(-\frac{1}{2(2n+1)}\) are left. Hence, the limit is the simplified expression \(\lim_{n \to \infty} \frac{1}{2} - \frac{1}{2(2n+1)}\).

Step 5: Evaluate the Limit

As \(n\) approaches infinity, the term \(\frac{1}{2(2n+1)}\) approaches 0, leaving us with \(\frac{1}{2}\). Therefore, the limit of the series is \(\frac{1}{2}\).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex fractions into simpler components. It is particularly useful in calculus for simplifying terms in a series or integral.

In this exercise, you are using partial fraction decomposition to rewrite each term of the series. The term given is \[ \frac{1}{(2k-1)(2k+1)}. \]
By applying partial fraction decomposition, you express it as a sum of simpler fractions:\[ \frac{A}{2k-1} + \frac{B}{2k+1}. \]
To find the values of \(A\) and \(B\), you need to solve the equation:\[ 1 = A(2k+1) + B(2k-1). \]
Upon solving this, you get \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\). Thus, the decomposition for each term becomes:\[ \frac{1}{2(2k-1)} - \frac{1}{2(2k+1)}. \]
This form is critical because it lays the groundwork for identifying the series as a telescoping series, where most terms will cancel each other out.

Calculating Limits

When you deal with sequences and series, calculating limits is a fundamental tool you use to understand the behavior as \(n\) approaches infinity.

In this problem, once you've rewritten the series using partial fraction decomposition and recognized its telescoping nature, you have:\[ \frac{1}{2} - \frac{1}{2(2n+1)}. \]
To find the limit as \(n\) approaches infinity, you evaluate the above expression. You note:

• The term \(\frac{1}{2(2n+1)}\) diminishes because \(2n+1\) grows larger and larger.
• The entire term tends to 0 as \(n\) increases without bound.

Therefore, what remains is just \(\frac{1}{2}\).

This demonstrates that the limit of the entire expression is \(\frac{1}{2}\). Calculating limits is crucial here because it allows us to understand the ultimate value of the stack of terms as "time goes on," and whether they converge or diverge.

Infinite Series

Infinite series involve sums of endlessly extending terms.
In calculus, infinite series can converge to a limit or diverge towards infinity or some undefined result.

The exercise gives the sum:\[ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n+1)(2n+3)}. \]
When dealing with an infinite series:

• It's essential to first check if it telescopes, as this makes it simpler to find its behavior.
• Telescoping means that intermediate terms cancel, leaving only the first and last terms.
• You often convert each term into a simpler form to see these cancellations.

In our case, the telescoping nature ensures a clearer insight:\[ \frac{1}{2} - \frac{1}{2(2n+1)}, \]
of which, once cancelled, only the remaining terms matter.

You find that for this infinite series as \(n\) tends towards infinity, the result stabilizes to \(\frac{1}{2}\), showcasing the beauty of convergence in infinite mathematics.