Problem 119 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Sum to infinite terms of the series \(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots .\) is (A) \(\frac{1}{4}\) (B) \(\frac{1}{3}\) (C) \(\frac{1}{2}\) (D) None of thesePassage 2 A general arithmetic progression is \(a, a+d, a+2 d, \ldots\) and a general geometric progression is \(a, a r, a r^{2}, \ldots\), then the sequence \(a,(a+d) r,(a+2 d) r^{2}, \ldots\) is called an arithmetico-geometric progression (A. G.P.). Note that each term of the A.G.P. is the product of the corresponding terms of the A.P. \(a, a+d, a+2 d, \ldots\) and the G.P. \(1, r, r^{2}, \ldots\) The \(n\)th term of the A.G.P. is \([a+(n-1) d] r^{n-1}\) Sum to \(n\) terms of A.G.P. Let \(S_{n}=a+(a+d) r+(a+2 d) r^{2}+\ldots\) $$ +(a+\overline{n-2} d) r^{n-2}+(a+\overline{n-1} d) r^{n-1} $$ \(\left.\Rightarrow \quad r S_{n}=a r+(a+d) r^{2}+\ldots+\mid a+\overline{n-1} d\right) r^{n}\) Subtracting (2) from (1), we get $$ \begin{aligned} &(1-r) S_{n}=a+d r+d r^{2}+\ldots+d r^{n-1}-(a+\overline{n-1} d) r^{n} \\ &=a+\left(d r+d r^{2}+\ldots+\text { to } \overline{n-1} \text { terms }\right)-(a+\overline{n-1} d) r^{n} \\ &\left.\quad=a+\frac{d r\left(1-r^{n-1}\right)}{1-r}-\mid a+\overline{n-1} d\right) r^{n} \\ &\therefore S_{n}=\frac{a}{1-r}+\frac{d r\left(1-r^{n-1}\right)}{(1-r)^{2}}-\frac{(a+\overline{n-1} d) r^{n}}{1-r} \end{aligned} $$ \(\therefore\) Sum to infinite terms \(=S=\lim _{n \rightarrow \infty} S_{n}=\frac{a}{1-r}+\frac{d r}{(1-r)^{2}}\),

Step-by-Step Solution

Answer

The sum to infinite terms of the series is \( \frac{1}{2} \), so the answer is (C).

Step 1: Identify the series terms

Identify the terms in the series \ \( \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots \). Notice that each term in the series is of the form \( \frac{1}{(2n-1)(2n+1)} \).

Step 2: Simplify a general term

Express each term \( \frac{1}{(2n-1)(2n+1)} \) as a difference of two fractions. Use partial fraction decomposition:\( \frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \).

Step 3: Write the series using partial fractions

Rewrite the series using the partial fraction form:\( \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots = \frac{1}{2} \left( 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \ldots \right) \).

Step 4: Recognize the telescoping series

Observe that the series is telescoping, as most intermediate terms cancel each other out:\( 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \ldots \) simplifies to a sequence where all terms cancel except for the first term of the series.

Step 5: Determine the sum to infinity

We see that the series effectively becomes:\( \frac{1}{2} \left( 1 \right) = \frac{1}{2} \).Thus, the sum to infinite terms of the series is \( \frac{1}{2} \).

Key Concepts

Arithmetic Progression

Arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is known as the "common difference."
The general form of an arithmetic progression (A.P.) can be expressed as:

• First term: \( a \)
• Second term: \( a + d \)
• Third term: \( a + 2d \)
• \[ \ldots \]
• \( n \)-th term: \( a + (n-1)d \)

Arithmetic progressions are straightforward as you simply need to keep adding the common difference to find subsequent terms.
For example, if the first term \( a \) is \( 2 \) and the common difference \( d \) is \( 3 \), then the sequence would be: \( 2, 5, 8, 11, \ldots \).
These progressions help us in finding or deriving sums of series, such as the sum of consecutive natural numbers.

Geometric Progression

Geometric progression is a type of sequence where each term after the first is found by multiplying the previous one by a fixed non-zero number called the "common ratio."
The general formula for a geometric progression (G.P.) is:

• First term: \( a \)
• Second term: \( ar \)
• Third term: \( ar^2 \)
• \[ \ldots \]
• \( n \)-th term: \( ar^{(n-1)} \)

A geometric progression becomes particularly useful when dealing with exponential growth or decay situations.
For instance, if the first term \( a \) is \( 3 \) and the common ratio \( r \) is \( 2 \), the sequence will be: \( 3, 6, 12, 24, \ldots \).
The sum of a finite G.P. can be calculated using the formula: \( S_n = a \frac{1-r^n}{1-r} \), where \( S_n \) is the sum of the first \( n \) terms. For an infinite G.P. where \(|r| < 1\), the sum formula changes to \( S = \frac{a}{1-r} \).

Telescoping Series

A telescoping series is a series where successive terms cancel each other out, simplifying the evaluation of the sum.
In a telescoping series, when you expand the terms, intermediate terms vanish, leaving just a few terms from the start and end.
This makes calculating the series much easier and is exactly what happens with the given series. Let's consider the given series:

• Each term is \( \frac{1}{(2n-1)(2n+1)} \)
• Using partial fraction decomposition, it becomes \( \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \)
• When expanded this turns into a telescoping form where terms cancel each other out.

In the solution, one could notice that after decomposing the terms, it simplifies to a sequence such as: \( 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \ldots \)
Most of the terms cancel, simplifying directly to just the series' first term, making it straightforward to determine the sum to infinity. Here, the given infinite series sum simplifies to \( \frac{1}{2} \).
Telescoping series are thus very powerful, especially for series with complex fraction products that can be split to reveal cancelling terms.