Odd Power Sum 65E09F

1. **State the problem:** We want to estimate the infinite series $$\sum_{n=0}^\infty \frac{1}{(2n+1)^9}$$ correct to five decimal places. 2. **Formula and approach:** This is a series over odd integers raised to the power 9. We can write it as $$\sum_{k=0}^\infty \frac{1}{(2k+1)^9}.$$ This series converges very fast because the exponent 9 is large. 3. **Relation to Riemann zeta function:** Recall the Riemann zeta function $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}.$$ We can express the sum over odd integers in terms of $$\zeta(9)$$: $$\sum_{k=0}^\infty \frac{1}{(2k+1)^9} = \left(1 - \frac{1}{2^9}\right) \zeta(9).$$ 4. **Calculate values:** - Calculate $$1 - \frac{1}{2^9} = 1 - \frac{1}{512} = \frac{511}{512} = 0.998046875.$$ - Use known value $$\zeta(9) \approx 1.0020083928260822.$$ 5. **Multiply:** $$\sum_{k=0}^\infty \frac{1}{(2k+1)^9} \approx 0.998046875 \times 1.0020083928260822 = 1.000041785.$$ 6. **Rounding:** To five decimal places, the sum is approximately $$1.00004.$$