calculate-n-1-17-1-n-n-3-

Question Number 75063 by mathmax by abdo last updated on 06/Dec/19

c a l c u l a t e \sum_{n = 1}^{17} \frac{{(- 1)}^{n}}{n^{3}}

Commented by mathmax by abdo last updated on 17/Dec/19

l e t A = \sum_{n . = 1}^{17} \frac{{(- 1)}^{n}}{n^{3}} \Rightarrow A = \sum_{p = 1}^{[\frac{17}{2}]} \frac{1}{{(2 p)}^{3}} - \sum_{p = 0}^{8} \frac{1}{{(2 p + 1)}^{3}}

= \frac{1}{8} \sum_{p = 1}^{8} \frac{1}{p^{3}} - \sum_{p = 0}^{8} \frac{1}{{(2 p + 1)}^{3}} a l s o

\sum_{p = 1}^{8} \frac{1}{p^{3}} = \sum_{p = 1}^{4} \frac{1}{{(2 p)}^{3}} + \sum_{p = 0}^{3} \frac{1}{{(2 p + 1)}^{3}} = \frac{1}{8} \sum_{p = 1}^{4} \frac{1}{p^{3}} + \sum_{p = 0}^{3} \frac{1}{{(2 p + 1)}^{3}}

\Rightarrow A = \frac{1}{8} {\frac{1}{8} \sum_{p = 1}^{4} \frac{1}{p^{3}} + \sum_{p = 0}^{3} \frac{1}{{(2 p + 1)}^{3}}} - \sum_{p = 0}^{8} \frac{1}{{(2 p + 1)}^{3}}

= \frac{1}{64} \sum_{p = 1}^{4} \frac{1}{p^{3}} + (\frac{1}{8} - 1) \sum_{p = 0}^{3} \frac{1}{{(2 p + 1)}^{3}} - \sum_{p = 4}^{8} \frac{1}{{(2 p + 1)}^{3}}

= \frac{1}{64} (1 + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \frac{1}{4^{3}}) - \frac{7}{8} (1 + \frac{1}{3^{3}} + \frac{1}{5^{3}} + \frac{1}{7^{3}})

- (\frac{1}{9^{3}} + \frac{1}{11^{3}} + \frac{1}{13^{3}} + \frac{1}{15^{3}} + \frac{1}{17^{3}}) = \dots .

Answered by tw000001 last updated on 08/Dec/19

Commented by tw000001 last updated on 08/Dec/19

I use Desmos to find the answer,

but {it}^{'} s impossible to use integral to solve,

so the answer is approximately at - 0.8241 .