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 (B) \(\frac{\pi^{4}}{96}\).
1Step 1: Understanding the Total Sum
The given series \( \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots \) converges to \( \frac{\pi^4}{90} \). This is a well-known result for the sum of the reciprocals of the fourth powers of natural numbers.
2Step 2: Separate Even and Odd Terms
The series can be split into even and odd components: \( \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \) and \( \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots \). We are interested in the sum of the odd terms.
3Step 3: Calculate the Sum of Even Terms
The sum of the even terms is \( \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots \). Factor out \( \frac{1}{2^4} \) to get: \[ \frac{1}{2^4}(1^{-4} + 2^{-4} + 3^{-4} + \cdots). \] Recognizing that this series is similar to the original series, it sums to \( \frac{\pi^4}{90} \), but halved: \[ \frac{1}{16} \times \frac{\pi^4}{90} = \frac{\pi^4}{1440}. \]
4Step 4: Find the Sum of Odd Terms
Subtract the sum of the even terms from the total series sum to obtain the sum of the odd terms: \[ \frac{\pi^4}{90} - \frac{\pi^4}{1440}. \] Simplifying this, the common denominator is 1440: \[ \frac{16\pi^4}{1440} - \frac{\pi^4}{1440} = \frac{15\pi^4}{1440} = \frac{\pi^4}{96}. \]
5Step 5: Conclusion
The value of the series of odd-numbered reciprocals, \( \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \), is computed to be \( \frac{\pi^4}{96} \).
Key Concepts
Zeta FunctionSeries ConvergenceReciprocal Powers
Zeta Function
The Zeta Function is a fascinating and important function in mathematics, especially in number theory and mathematical analysis. It comes up in various areas, particularly in the study of series related to powers of integers. The most famous version of it is the Riemann Zeta function, denoted as \( \zeta(s) \), where
The series is all about summing the reciprocals of the integers raised to the power of \( s \). The function has some mind-boggling applications, including connections to prime numbers and the famous unsolved Riemann Hypothesis.
In the context of the problem, we deal with the specific case \( s = 4 \), which is \( \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} \). This is a special value and part of the broader family of zeta function evaluations known as Apéry's constants.
- \( s \) is a complex number.
- For values of \( s > 1 \), it is defined as \( \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \).
The series is all about summing the reciprocals of the integers raised to the power of \( s \). The function has some mind-boggling applications, including connections to prime numbers and the famous unsolved Riemann Hypothesis.
In the context of the problem, we deal with the specific case \( s = 4 \), which is \( \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} \). This is a special value and part of the broader family of zeta function evaluations known as Apéry's constants.
Series Convergence
Series Convergence is the idea that an infinite sum can yield a finite number. This might seem strange at first because we often think of infinity as an unreachable destination. However, in mathematics, series can indeed be finite if they converge.
Convergence happens when the terms of a series decrease rapidly enough as the series progresses. For example, in the series \( \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots \), the terms get smaller quickly because higher powers grow fast.
We determine if a series converges in various ways. A popular method is the comparison test, where one matches the behavior of a complex series with a simpler, known converging series. When the terms of our series decrease similarly or faster, convergence can be inferred.
Convergence happens when the terms of a series decrease rapidly enough as the series progresses. For example, in the series \( \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots \), the terms get smaller quickly because higher powers grow fast.
We determine if a series converges in various ways. A popular method is the comparison test, where one matches the behavior of a complex series with a simpler, known converging series. When the terms of our series decrease similarly or faster, convergence can be inferred.
- The problem's total sum is given: \( \frac{\pi^4}{90} \).
- This result illustrates convergence because the infinite sum reaches a finite known value.
Reciprocal Powers
Reciprocal Powers involve working with fractions that have powers of natural numbers in their denominators. They are intuitive yet powerful constructs deeply involved in series analysis.
They typically look like \( \frac{1}{n^a} \), where \( n \) is a natural number and \( a \) is a positive integer. These fractions decrease quickly as \( n \) increases, especially when \( a \) is large.
The series used in the exercise, \( \frac{1}{1^4}, \frac{1}{2^4}, \frac{1}{3^4}, \ldots \), is an illustrative example where \( a = 4 \). Such constructions lead to sums that converge remarkably fast.
Understanding reciprocal powers is vital in analyzing sum behaviors and determining convergence across mathematical tasks.
They typically look like \( \frac{1}{n^a} \), where \( n \) is a natural number and \( a \) is a positive integer. These fractions decrease quickly as \( n \) increases, especially when \( a \) is large.
The series used in the exercise, \( \frac{1}{1^4}, \frac{1}{2^4}, \frac{1}{3^4}, \ldots \), is an illustrative example where \( a = 4 \). Such constructions lead to sums that converge remarkably fast.
- In the problem, the reciprocal terms split by odd and even numbers to find specific summed values.
- This splitting is a common technique to handle complicated series by simplifying some parts for easier calculation.
Understanding reciprocal powers is vital in analyzing sum behaviors and determining convergence across mathematical tasks.
Other exercises in this chapter
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 87
In a \(\Delta \mathrm{abc}\), if \(\cot A, \cot B, \cot C\) are in A.P. then \(a^{2}, b^{2}\), \(c^{2}\) are in (A) A.P. (B) G.P. (C) H.P. (D) A.G. P.
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