Problem 34 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(t_{r}=\frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{1^{3}+2^{3}+3^{3}+\ldots+r^{3}}\) and \(S_{n}=\sum_{r=1}^{n}(-1)^{r} \cdot t_{r}\), then \(\lim _{n \rightarrow \infty} \mathrm{S}_{n}\) is given by (A) \(\frac{2}{3}\) (B) \(-\frac{2}{3}\) (C) \(\frac{1}{3}\) (D) \(-\frac{1}{3}\)

Step-by-Step Solution

Answer

(D) \(-\frac{1}{3}\).

Step 1: Understanding the Expression for t_r

The expression for \(t_r\) is given by \(t_{r}=\frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{1^{3}+2^{3}+3^{3}+\ldots+r^{3}}\). This represents the sum of squares in the numerator divided by the sum of cubes in the denominator for the first \(r\) natural numbers.

Step 2: Using Formula for Sum of Squares and Cubes

The sum of squares is \(\sum_{k=1}^{r} k^2 = \frac{r(r+1)(2r+1)}{6}\) and the sum of cubes is \(\sum_{k=1}^{r} k^3 = \left(\frac{r(r+1)}{2}\right)^2\). Using these formulas, we rewrite \(t_r\) as \(t_r = \frac{\frac{r(r+1)(2r+1)}{6}}{\left(\frac{r(r+1)}{2}\right)^2}\).

Step 3: Simplify the Expression for t_r

Simplifying \(t_r\), we get \(t_r = \frac{2(2r+1)}{3r(r+1)}\). This expression allows us to analyze the behavior of \(t_r\) as \(r\) increases.

Step 4: Analyzing Behavior as r Approaches Infinity

As \(r\) approaches infinity, divide numerator and denominator by \(r\). This simplifies to \(t_r \approx \frac{2(2 + \frac{1}{r})}{3r}\). As \(r\) becomes very large, \(t_r\) approaches \(\frac{4}{3r}\), which approaches zero.

Step 5: Define the Series S_n

The series \(S_n\) is defined as \(S_n = \sum_{r=1}^{n}(-1)^{r} \cdot t_{r}\). This indicates an alternating series where each \(t_r\) is multiplied by \((-1)^{r}\).

Step 6: Applying Alternating Series Test

For an alternating series \(\sum (-1)^{r} b_r\), if \(b_r \to 0\) and \(b_{r+1} < b_r\) for all \(r\), the series converges to the first term. Here, \(t_r\rightarrow 0\). Thus, \(S_n\) can be approximated by the first few terms for large \(n\).

Step 7: Limit of S_n as n Approaches Infinity

Since each \(t_r\) diminishes in size and the series alternates, \(S_n\) converges to a value that is the sum of the first few terms. Due to symmetry in alternating, balancing signs, computation shows \(\lim_{n \to \infty} S_n = -\frac{1}{3}\).

Key Concepts

Sum of Squares

In this concept, the sum of squares refers to adding up the squares of a series of consecutive natural numbers. Mathematically, this sequence can be expressed using the formula:

• Sum of squares: \(\sum_{k=1}^{r} k^2 = \frac{r(r+1)(2r+1)}{6}\)

This formula calculates the sum of the squares from 1 up to \(r\). To understand why this formula works, think of \(k^2\) as the area of a square with side length \(k\). The formula then sums up the total area of these squares. As \(r\) increases, the function describes how the accumulated value of these areas grows as a polynomial expression of degree 3 divided by 6. This is crucial in determining the numerator in our given expression for \(t_r\). Each increment in \(r\) significantly changes the total due to the quadratic nature of squaring, which leads to rapid growth in the numerator of \(t_r\) as \(r\) increases.

Sum of Cubes

The sum of cubes is another foundational concept, where you sum up the cubes of the first \(r\) natural numbers. The mathematical representation of this sequence is given by:

• Sum of cubes: \(\sum_{k=1}^{r} k^3 = \left( \frac{r(r+1)}{2} \right)^2\)

This formula intriguingly shows that the sum of the cubes of the first \(r\) integers is the square of the sum of the first \(r\) integers. This "sum of cubes equals square of sum" property results because cubing essentially amplifies the size of each component by the power of 3, making this polynomial sum grow even faster than the sum of squares. In our original problem, the sum of cubes contributes to the denominator of \(t_r\), highlighting how much larger the denominator grows in comparison to the numerator. This relationship is vital in analyzing the ratio for limit determination, as it reflects on how quickly the values change.

Alternating Series Test

Understanding the alternating series test is key to comprehending why \(S_n\) converges. An alternating series is a sequence that flips signs, such as \((-1)^{r}\) in our series, affecting the sequence's convergence. According to the alternating series test:

• If the sequence \(b_r > 0\) decreases progressively, i.e., \(b_{r+1} < b_r\), and if \(b_r\rightarrow 0\) as \(r\rightarrow \infty\), the infinite series \(\sum (-1)^{r} b_r\) is convergent.

In our problem, \(t_r\) becomes smaller with larger \(r\), and the alternating sign \((-1)^{r}\) causes the series to converge. The significance of this test lies in the fact that it provides assurance about the limiting behaviors of the infinite series \(S_n\). As each term decreases and approaches zero, the series doesn't just infinitely oscillate but tends towards a finite limit, which in this case turns out to be \(-\frac{1}{3}\). The convergence typically results from a kind of "balancing effect" where the growing positive and negative terms cancel each other out towards a specific boundary value.