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What is \(\frac{3}{1^2 \times 2^2} + \frac{5}{2^2 \times 3^2} + \frac{7}{3^2 \times 4^2} + ...\) equal to ?

This question was previously asked in
CDS 2 2024 Maths Question Paper (01-Sep-2024)
The correct answer is
1

To solve the problem of determining the sum \(\frac{3}{1^2 \times 2^2} + \frac{5}{2^2 \times 3^2} + \frac{7}{3^2 \times 4^2} + \ldots\), we first need to look for a pattern or a general term in the series.

Let's examine the terms clearly:

  • The numerator seems to increase by 2 each time: \(3, 5, 7, \ldots\)
  • The denominator is a product of consecutive squares, which can be represented as \(n^2 \times (n+1)^2\).

Given this analysis, the general term of the series can be expressed as:

\(\frac{2n + 1}{n^2 \times (n+1)^2}\)

Now, let's reconsider the series using partial fraction decomposition, which is often a useful strategy for managing such sequences:

The expression \(\frac{2n + 1}{n^2(n+1)^2}\) can be written as a difference of two terms:

\( \frac{1}{n^2} - \frac{1}{(n+1)^2} \)

This decomposition suggests a telescopic series, where most terms cancel out except the first and last ones as \(n\) approaches infinity. Let's write out a few terms:

  1. For n=1:

\(\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}\)

  1. For n=2:

\(\frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}\)

Adding these and continuing to higher \(n\), we observe that the series collapses, leaving:

\( 1 - \lim_{n \to \infty} \frac{1}{(n+1)^2} = 1 \)

Thus, the value of the infinite sum is indeed 1.

Therefore, the correct answer is 1.

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