Question Number 75063 by mathmax by abdo last updated on 06/Dec/19
$${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\mathrm{17}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} } \\ $$
Commented by mathmax by abdo last updated on 17/Dec/19
![let A =Σ_(n.=1) ^(17) (((−1)^n )/n^3 ) ⇒ A =Σ_(p=1) ^([((17)/2)]) (1/((2p)^3 )) −Σ_(p=0) ^8 (1/((2p+1)^3 )) =(1/8)Σ_(p=1) ^8 (1/p^3 )−Σ_(p=0) ^(8 ) (1/((2p+1)^3 )) also Σ_(p=1) ^8 (1/p^3 ) =Σ_(p=1) ^4 (1/((2p)^3 )) +Σ_(p=0) ^3 (1/((2p+1)^3 )) =(1/8)Σ_(p=1) ^4 (1/p^3 ) +Σ_(p=0) ^3 (1/((2p+1)^3 )) ⇒A =(1/8){(1/8) Σ_(p=1) ^4 (1/p^3 ) +Σ_(p=0) ^3 (1/((2p+1)^3 ))}−Σ_(p=0) ^8 (1/((2p+1)^3 )) =(1/(64)) Σ_(p=1) ^4 (1/p^3 ) +((1/8)−1)Σ_(p=0) ^3 (1/((2p+1)^3 ))−Σ_(p=4) ^8 (1/((2p+1)^3 )) =(1/(64))( 1+(1/2^3 )+(1/3^3 )+(1/4^3 ))−(7/8)(1+(1/3^3 ) +(1/5^3 ) +(1/7^3 )) −((1/9^3 )+(1/(11^3 )) +(1/(13^3 )) +(1/(15^3 )) +(1/(17^3 ))) =....](https://www.tinkutara.com/question/Q75813.png)
$${let}\:{A}\:=\sum_{{n}.=\mathrm{1}} ^{\mathrm{17}} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} }\:\Rightarrow\:{A}\:=\sum_{{p}=\mathrm{1}} ^{\left[\frac{\mathrm{17}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{3}} }\:−\sum_{{p}=\mathrm{0}} ^{\mathrm{8}} \frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\sum_{{p}=\mathrm{1}} ^{\mathrm{8}} \frac{\mathrm{1}}{{p}^{\mathrm{3}} }−\sum_{{p}=\mathrm{0}} ^{\mathrm{8}\:} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{3}} }\:{also} \\ $$$$\sum_{{p}=\mathrm{1}} ^{\mathrm{8}} \frac{\mathrm{1}}{{p}^{\mathrm{3}} }\:=\sum_{{p}=\mathrm{1}} ^{\mathrm{4}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{3}} }\:+\sum_{{p}=\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{8}}\sum_{{p}=\mathrm{1}} ^{\mathrm{4}} \:\frac{\mathrm{1}}{{p}^{\mathrm{3}} }\:+\sum_{{p}=\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow{A}\:=\frac{\mathrm{1}}{\mathrm{8}}\left\{\frac{\mathrm{1}}{\mathrm{8}}\:\sum_{{p}=\mathrm{1}} ^{\mathrm{4}} \:\frac{\mathrm{1}}{{p}^{\mathrm{3}} }\:+\sum_{{p}=\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{3}} }\right\}−\sum_{{p}=\mathrm{0}} ^{\mathrm{8}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\:\sum_{{p}=\mathrm{1}} ^{\mathrm{4}} \:\frac{\mathrm{1}}{{p}^{\mathrm{3}} }\:+\left(\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{1}\right)\sum_{{p}=\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{3}} }−\sum_{{p}=\mathrm{4}} ^{\mathrm{8}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\left(\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }\right)−\frac{\mathrm{7}}{\mathrm{8}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }\right) \\ $$$$−\left(\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{11}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{13}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{15}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{17}^{\mathrm{3}} }\right)\:=…. \\ $$
Answered by tw000001 last updated on 08/Dec/19
Commented by tw000001 last updated on 08/Dec/19
$$\mathrm{I}\:\mathrm{use}\:\mathrm{Desmos}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{answer}, \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{use}\:\mathrm{integral}\:\mathrm{to}\:\mathrm{solve}, \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{approximately}\:\mathrm{at}\:−\mathrm{0}.\mathrm{8241}. \\ $$