Math Problem Statement

If 1 ^ - 3 + 2 ^ - 3 + 3 ^ - 3 +...... upto∞ terms = p and 1 ^ - 3

  • 3 ^ - 3 + 5 ^ - 3 +... upto∞ terms = q , then p/q is a/b (where a, b are positive co-prime integers), then

a + b is

(1) 32

(2) 16

(3) 15

(4) 27

Solution

To solve this problem, we need to find the ratio pq\frac{p}{q} where

  • p=13+23+33+p = 1^{-3} + 2^{-3} + 3^{-3} + \cdots (sum of reciprocals of cubes of all positive integers)
  • q=13+33+53+q = 1^{-3} + 3^{-3} + 5^{-3} + \cdots (sum of reciprocals of cubes of all positive odd integers)

Step-by-Step Solution:

  1. Understanding pp:

    pp is the Riemann zeta function at s=3s = 3, which is known as ζ(3)\zeta(3): p=ζ(3)=n=11n3p = \zeta(3) = \sum_{n=1}^{\infty} \frac{1}{n^3}

  2. Understanding qq:

    qq is the sum of reciprocals of cubes of all positive odd integers: q=13+33+53+=n=01(2n+1)3q = 1^{-3} + 3^{-3} + 5^{-3} + \cdots = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^3}

  3. Expressing pp and qq:

    To find a relationship between pp and qq, consider separating the terms of pp into sums over odd and even integers: p=n=11n3=odd n1n3+even n1n3p = \sum_{n=1}^{\infty} \frac{1}{n^3} = \sum_{\text{odd } n} \frac{1}{n^3} + \sum_{\text{even } n} \frac{1}{n^3} Let's denote the sum over odd integers as qq, and the sum over even integers separately: even n1n3=m=11(2m)3=m=118m3=18m=11m3=ζ(3)8\sum_{\text{even } n} \frac{1}{n^3} = \sum_{m=1}^{\infty} \frac{1}{(2m)^3} = \sum_{m=1}^{\infty} \frac{1}{8m^3} = \frac{1}{8} \sum_{m=1}^{\infty} \frac{1}{m^3} = \frac{\zeta(3)}{8} Therefore, p=q+ζ(3)8p = q + \frac{\zeta(3)}{8}

  4. Finding pq\frac{p}{q}:

    Using p=ζ(3)p = \zeta(3) and the relation p=q+ζ(3)8p = q + \frac{\zeta(3)}{8}: ζ(3)=q+ζ(3)8\zeta(3) = q + \frac{\zeta(3)}{8} q=ζ(3)ζ(3)8=8ζ(3)ζ(3)8=7ζ(3)8q = \zeta(3) - \frac{\zeta(3)}{8} = \frac{8\zeta(3) - \zeta(3)}{8} = \frac{7\zeta(3)}{8} Thus, pq=ζ(3)7ζ(3)8=ζ(3)×87ζ(3)=87\frac{p}{q} = \frac{\zeta(3)}{\frac{7\zeta(3)}{8}} = \frac{\zeta(3) \times 8}{7\zeta(3)} = \frac{8}{7}

  5. Finding a+ba + b:

    Since pq=87\frac{p}{q} = \frac{8}{7}, a=8a = 8 and b=7b = 7. Therefore, a+b=8+7=15a + b = 8 + 7 = 15.

Conclusion

The correct answer is:

(3) 15

Would you like further details on any step? Here are five related questions you might find interesting:

  1. What is the Riemann zeta function, and why is it significant in mathematics?
  2. Can you explain more about the convergence of series like n=11n3\sum_{n=1}^{\infty} \frac{1}{n^3}?
  3. How do properties of odd and even numbers affect summation in series?
  4. Are there similar problems that involve other values of nn for the Riemann zeta function?
  5. What are some applications of the Riemann zeta function in physics or number theory?

Tip: When dealing with infinite series, breaking them into smaller components often helps in finding relationships between them.

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Math Problem Analysis

Mathematical Concepts

Infinite series
Riemann zeta function

Formulas

\( \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \)

Theorems

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Suitable Grade Level

Advanced Mathematics