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

$$calculate\sum _{n=1}^{17}\frac{{(-1)}^n}{n^3}$$

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

$$letA=\sum _{n.=1}^{17}\frac{{(-1)}^n}{n^3}\Rightarrow A=\sum _{p=1}^{\left[\frac{17}{2}\right]}\frac{1}{{\left(2p\right)}^3}-\sum _{p=0}^8\frac{1}{{(2p+1)}^3}$$$$=\frac{1}{8}\sum _{p=1}^8\frac{1}{p^3}-\sum _{p=0}^8\frac{1}{{(2p+1)}^3}also$$$$\sum _{p=1}^8\frac{1}{p^3}=\sum _{p=1}^4\frac{1}{{\left(2p\right)}^3}+\sum _{p=0}^3\frac{1}{{(2p+1)}^3}=\frac{1}{8}\sum _{p=1}^4\frac{1}{p^3}+\sum _{p=0}^3\frac{1}{{(2p+1)}^3}$$$$\Rightarrow A=\frac{1}{8}\{\frac{1}{8}\sum _{p=1}^4\frac{1}{p^3}+\sum _{p=0}^3\frac{1}{{(2p+1)}^3}\}-\sum _{p=0}^8\frac{1}{{(2p+1)}^3}$$$$=\frac{1}{64}\sum _{p=1}^4\frac{1}{p^3}+(\frac{1}{8}-1)\sum _{p=0}^3\frac{1}{{(2p+1)}^3}-\sum _{p=4}^8\frac{1}{{(2p+1)}^3}$$$$=\frac{1}{64}(1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3})-\frac{7}{8}(1+\frac{1}{3^3}+\frac{1}{5^3}+\frac{1}{7^3})$$$$-(\frac{1}{9^3}+\frac{1}{{11}^3}+\frac{1}{{13}^3}+\frac{1}{{15}^3}+\frac{1}{{17}^3})=....$$

Answered by tw000001 last updated on 08/Dec/19

Commented by tw000001 last updated on 08/Dec/19

$$\mathrm{I}\mathrm{use}\mathrm{Desmos}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{answer},$$$$\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{impossible}\mathrm{to}\mathrm{use}\mathrm{integral}\mathrm{to}\mathrm{solve},$$$$\mathrm{so}\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{approximately}\mathrm{at}-0.8241.$$

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