Problem 13 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\lambda=\sum_{i=1}^{\infty} \frac{1}{i^{4}}\), then \(\sum_{i=1}^{\infty} \frac{1}{(2 i-1)^{4}}\) is (A) \(\frac{14}{15} \lambda\) (B) \(\frac{\lambda}{2}\) (C) \(\frac{16}{15} \lambda\) (D) \(\frac{15}{16} \lambda\)

Step-by-Step Solution

Answer

The answer is option (D) \(\frac{15}{16} \lambda\).

Step 1: Understanding the Problem

We need to find the value of the series \(\sum_{i=1}^{\infty} \frac{1}{(2i-1)^4}\), given that \(\lambda = \sum_{i=1}^{\infty} \frac{1}{i^4}\). We will relate this to the given \(\lambda\).

Step 2: Breaking Down the Series

Consider \(\lambda\) which is the whole series \(\sum_{i=1}^{\infty} \frac{1}{i^4}\). It can be split into two parts: the series of terms where \(i\) is odd and the series of terms where \(i\) is even.

Step 3: Odd and Even Terms

For odd \(i\), the terms become \(\frac{1}{(2i-1)^4}\) and for even \(i\), the terms become \(\frac{1}{(2i)^4} = \frac{1}{16i^4}\).

Step 4: Expressing \(\lambda\) as Two Series

Thus, \(\lambda = \left( \sum_{i=1}^{\infty} \frac{1}{(2i-1)^4} \right) + \left( \sum_{i=1}^{\infty} \frac{1}{16i^4} \right)\).

Step 5: Relate to Known Series

Let \(S = \sum_{i=1}^{\infty} \frac{1}{(2i-1)^4}\). So we have \(\lambda = S + \frac{1}{16} \cdot \lambda\), because the even series represents \(\frac{1}{16}\) of \(\lambda\).

Step 6: Solving for S

Rearrange: \(\lambda - \frac{1}{16}\lambda = S\). This simplifies to \(\frac{15}{16}\lambda = S\). Thus, \(S = \frac{15}{16} \lambda\).

Key Concepts

Convergence of Series

Understanding the convergence of a series is crucial when dealing with infinite series. An infinite series is essentially a sequence of partial sums. The big question in convergence is whether these sums grow without bound or approach a finite number.
In the problem given, the series converges because the terms \( \frac{1}{i^4} \) decrease rapidly as \( i \) increases. This series is known to converge to a specific finite value, \( \lambda \).

• A series converges if its terms approach zero quickly enough.
• For a series \( \sum a_i \) to converge, the sequence of partial sums must approach a finite limit.

Recognizing convergence or divergence helps in deciphering and manipulating these series into meaningful mathematical results.

Mathematical Problem Solving

Effective mathematical problem solving involves a combination of understanding, strategy, and execution.
The first step in tackling problems like the one given is to grasp clearly what you are asked to find and understand how given information, such as \( \lambda \), fits into the problem.
Breaking the problem into smaller, manageable parts is a powerful technique. Here, we split the series \( \sum \frac{1}{i^4} \) into odd and even terms, making the problem tractable.

• Odd terms are expressed as \( \frac{1}{(2i-1)^4} \).
• Even terms are expressed as \( \frac{1}{16i^4} \).

This breakdown leads to a strategic rearrangement of the equation to isolate \( S \), where this skillful decomposition plays a pivotal role in successfully solving the problem.

Harmonic Series

The harmonic series is a well-known divergent series given by \( \sum \frac{1}{i} \).
Unlike the convergent series we explored in the exercise, the harmonic series increases without bound as more terms are added, which means it grows indefinitely.
Understanding the harmonic series provides a baseline for comparing behaviors of other similar series in mathematical analysis. This knowledge helps you recognize why specific manipulations, such as splitting into odd and even terms, work effectively with the convergent series initially provided.

• The harmonic series demonstrates that not all series with terms approaching zero will converge.
• This series is crucial in understanding the conditions needed for convergence versus divergence.

While the harmonic series itself does not directly apply to solving the given problem, knowing about it enriches your broader understanding of series convergence.