Problem 88
Question
If \(\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots .\) up to \(\infty=\frac{\pi^{4}}{90}\), then the value of \(\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\ldots .\) up to \(\infty\) is (A) \(\frac{\pi^{4}}{45}\) (B) \(\frac{\pi^{4}}{96}\) (C) \(\frac{\pi^{4}}{124}\) (D) None of these
Step-by-Step Solution
Verified Answer
The value of the series is \( \frac{\pi^4}{96} \) (Option B).
1Step 1: Understand the Given Series
The given series \( \frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots \) up to infinity equals \( \frac{\pi^{4}}{90} \). This is a known result for the Riemann zeta function \( \zeta(4) \). It reveals that the sum of the reciprocals of the fourth powers of all natural numbers equals \( \frac{\pi^{4}}{90} \).
2Step 2: Identify the New Series
The problem asks for the sum of \( \frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + \ldots \) up to infinity. This series involves only the reciprocals of the fourth powers of odd integers.
3Step 3: Express the Series
The original series \( \zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} \) can be split into two parts: the sum of reciprocals of the fourth powers of odd and even integers. Let \( S_1 = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \) and \( S_2 = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots \).
4Step 4: Calculate the Even Series Contribution
Notice that \( S_2 \) can be written as \( \sum_{k=1}^{\infty} \frac{1}{(2k)^4} = \frac{1}{16} \sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{1}{16} \zeta(4) \). So, \( S_2 = \frac{\pi^4}{1440} \).
5Step 5: Find the Odd Series Contribution
Since \( \zeta(4) = S_1 + S_2 \) and \( \zeta(4) = \frac{\pi^4}{90} \), we have \( \frac{\pi^4}{90} = S_1 + \frac{\pi^4}{1440} \). Solving for \( S_1 \), we find \( S_1 = \frac{\pi^4}{90} - \frac{\pi^4}{1440} = \frac{\pi^4}{96} \).
6Step 6: Verify and Conclude
We computed \( S_1 = \frac{\pi^4}{96} \), which matches one of the provided options. Therefore, option (B) is correct.
Key Concepts
Series ConvergenceRiemann Zeta Function IdentitiesEven and Odd Series
Series Convergence
Series convergence is a fundamental concept in mathematics that deals with the behavior of infinitely long sums. When a series converges, it means that as you keep adding its terms, the sum approaches a fixed number. This is particularly important in mathematical analysis and number theory because many functions and phenomena can be expressed as infinite series.
To determine if a series converges, we often use tests like the Ratio Test, the Root Test, or comparison tests against known convergent series. For example, the series \( \frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \ldots \) up to infinity converges to \( \frac{\pi^4}{90} \), a result connected to the Riemann zeta function. This convergence occurs because the terms decrease rapidly as the powers increase, leading to a finite sum despite the infinite number of terms.
To determine if a series converges, we often use tests like the Ratio Test, the Root Test, or comparison tests against known convergent series. For example, the series \( \frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \ldots \) up to infinity converges to \( \frac{\pi^4}{90} \), a result connected to the Riemann zeta function. This convergence occurs because the terms decrease rapidly as the powers increase, leading to a finite sum despite the infinite number of terms.
- A series is convergent if its terms get smaller and approach zero.
- The sum of a convergent series is called its limit.
- In the series \( \frac{1}{n^4} \), the decrement by the power of 4 ensures convergence.
Riemann Zeta Function Identities
The Riemann zeta function, denoted by \( \zeta(s) \), is a function of complex numbers that analytically continues the sum of the reciprocals of the natural numbers raised to a power \( s \), expressed as \( \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \). This function is profoundly significant in number theory, particularly in understanding the distribution of prime numbers. In the given exercise, we specifically used \( \zeta(4) \) which directly refers to the series \( \frac{1}{1^4} + \frac{1}{2^4} + \ldots \).
The zeta function has several fascinating identities and properties:
The zeta function has several fascinating identities and properties:
- For even positive integers, there are explicit formulas involving powers of \( \pi \).
- \( \zeta(2) = \frac{\pi^2}{6} \), and \( \zeta(4) = \frac{\pi^4}{90} \).
- These identities help solve problems by relating complex sums to simpler, known values.
Even and Odd Series
When dealing with series, one effective strategy is to consider the sum of even-indexed and odd-indexed terms separately. This strategy is particularly useful when the terms of the series behave differently for even and odd indices.
In the given exercise, the series \( \zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} \) can be split into two parts: the sum of reciprocals of the fourth powers of odd numbers and even numbers. The even-numbered series (\( S_2 \)) was calculated first, demonstrating how the manipulation of series terms for specific indices can simplify complex calculations.
In the given exercise, the series \( \zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} \) can be split into two parts: the sum of reciprocals of the fourth powers of odd numbers and even numbers. The even-numbered series (\( S_2 \)) was calculated first, demonstrating how the manipulation of series terms for specific indices can simplify complex calculations.
- Odd terms: \( \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \)
- Even terms scaled by fourth powers: \( \frac{1}{2^4} + \frac{1}{4^4} + \ldots \) can be rewritten as \( \frac{1}{16}\sum_{n=1}^{\infty} \frac{1}{n^4} \).
- This decomposition results in two simpler series that we combine to find the original sum.
Other exercises in this chapter
Problem 85
If, in a G.P. of \(3 n\) terms, \(S_{1}\) denotes the sum of the first \(n\) terms, \(S_{2}\) the sum of the second block of \(n\) terms and \(S_{3}\) the sum o
View solution Problem 86
In a geometric series, the first term is \(a\) and common ratio is \(r\). If \(\mathrm{S}_{n}\) denotes the sum of \(n\) terms and \(U_{n}\) \(=\sum_{n=1}^{n} \
View solution Problem 89
If the \((m+1)\) th, \((n+1)\) th and \((r+1)\) th terms of an A.P. are in G.P. and \(m, n, r\) are in H.P., then the ratio of the first term of the A.P. to its
View solution Problem 90
Let there be \(n\) numbers in G.P. whose common ratio is \(r\) and \(S_{m}\) denotes the sum of their first \(m\) terms. The sum of their products taken two at
View solution