Question

By grouping and rearranging terms of the absolutely convergent series $$ S=\sum_{n=1}^{\infty} \frac{1}{n^{2}} $$ show that $$ S_{\mathrm{o}}=\sum_{n \text { odd }}^{\infty} \frac{1}{n^{2}}=\frac{3 S}{4} $$

Short answer

\( S_{o} = \frac{3S}{4} \)

Step-by-step solution

Identify the Series

The given series is an absolutely convergent series: \[ S = \sum_{n=1}^{\infty} \frac{1}{n^2} \]

Separate Odd and Even Terms

Rewrite the series by separating the terms where `n` is odd and where `n` is even.\[ S = \sum_{n \text { odd }}^{\infty} \frac{1}{n^2} + \sum_{n \text { even }}^{\infty} \frac{1}{n^2} \]

Express Series for Even Terms

Note that for even terms, \(n\) can be written as \(2k\). Substitute and simplify:\[ \sum_{n \text { even }}^{\infty} \frac{1}{n^2} = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} = \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2} \]

Relation Between Even and Original Series

Recognize that \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) is the original series \(S\). Hence:\[ \sum_{n \text { even }}^{\infty} \frac{1}{n^2} = \frac{1}{4} S \]

Express Odd Terms Series

Using that \(S\) is separated into odd and even terms:\[ S = S_{o} + \frac{1}{4} S \]

Solve for \(S_{o}\)

Rearrange and solve for the series of odd terms \(S_{o}\):\[ S_{o} = S - \frac{1}{4} S = \frac{3S}{4} \]

Key concepts

Separating Series Terms

When dealing with series, it's often useful to separate terms to better understand or simplify the problem. In our case, the series is: \( S = \sum_{n=1}^{\infty} \frac{1}{n^2} \) By separating terms based on a pattern of even and odd indices, we can gain insights into the structure of the series. For the given series, we can rewrite it in terms of odd and even indices: \( S = \sum_{n \text { odd }}^{\infty} \frac{1}{n^2} + \sum_{n \text { even }}^{\infty} \frac{1}{n^2} \) This approach helps by isolating parts of the series that may reveal hidden relationships or simplifications. Separating terms is a powerful tool in series analysis because it can transform a complex series into more manageable parts.

Even and Odd Terms in Series

Separating a series into its even and odd terms can sometimes uncover elegant patterns and relationships. For even terms in our series, we observe the following: Let \( n \) be even. Then \( n = 2k \), where \( k \) is an integer. Rewriting the even terms we get: \[ \sum_{n \text { even }}^{\infty} \frac{1}{n^2} = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} \] This simplifies further to: \[ \sum_{k=1}^{\infty} \frac{1}{(2k)^2} = \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2} \] Notice here that \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) is precisely the original series \( S \). Therefore: \[ \sum_{n \text { even }}^{\infty} \frac{1}{n^2} = \frac{1}{4} S \] The odd terms, then, involve taking the difference between the total series \( S \) and the sum of the even terms: \[ S = S_{o} + \frac{1}{4} S \] Rearranging this, it becomes clear that: \[ S_{o} = S - \frac{1}{4} S = \frac{3S}{4} \]

Mathematical Approximation

Mathematical approximation comes into play when we simplify expressions or series to understand patterns or results more clearly. In the given problem, we approximate and simplify parts of the series. The initial series is broken down with approximations. To approximate the even term series, we used: \( \sum_{n \text { even }}^{\infty} \frac{1}{n^2} = \frac{1}{4} S \) This is achieved by understanding that for an even index, the series behaves similarly to the original but scaled down by a factor. Also, note that isolating the odd terms helps in revealing other approximations. For the odd term series, we generated: \( S_{o} = S - \frac{1}{4} S = \frac{3S}{4} \) This approximation captures the essence without overly complicating the original problem. Approximation is essential as it helps us understand complex series by breaking them into simpler forms and revealing intrinsic properties.