Question Number 210515 by hardmath last updated on 11/Aug/24
$$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{11}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{13}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{17}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{19}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{23}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{29}^{\mathrm{2}} }\:+\:…\:=\:? \\ $$
Commented by mr W last updated on 11/Aug/24
$${please}\:{recheck}! \\ $$$${is}\:{the}\:{last}\:{given}\:{term}\:\frac{\mathrm{1}}{\mathrm{25}^{\mathrm{2}} }\:{or}\:{really}\:\frac{\mathrm{1}}{\mathrm{29}^{\mathrm{2}} }? \\ $$
Commented by hardmath last updated on 11/Aug/24
$$ \\ $$Dear professor, the condition is given in this way
Commented by hardmath last updated on 11/Aug/24
$$ \\ $$Dear professpr, maybe there was a mistake
Commented by mr W last updated on 11/Aug/24
$${when}\:{the}\:{question}\:{is}\:{as}\:{you}\:{wrote}, \\ $$$${then}\:{it}\:{has}\:{no}\:{sense},\:{not}\:{to}\:{say}\:{that} \\ $$$${we}\:{can}\:{solve}\:{it}. \\ $$
Commented by hardmath last updated on 11/Aug/24
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Commented by aleks041103 last updated on 11/Aug/24
$${Maybe}\:{the}\:{demominators}\:{are}\:{the} \\ $$$${squares}\:{of}\:{primes}. \\ $$
Commented by mr W last updated on 11/Aug/24
$${maybe}.\:{but}\:{then}\:{he}\:{could}\:{have}\:{made} \\ $$$${clear}\:{with} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }+{etc}.+\frac{\mathrm{1}}{\mathrm{29}^{\mathrm{2}} }+… \\ $$$${but}\:{even}\:{though}\:{can}\:{we}\:{solve} \\ $$$$\underset{{all}\:{primes}\:} {\sum}\frac{\mathrm{1}}{{p}^{\mathrm{2}} }=? \\ $$
Commented by aleks041103 last updated on 11/Aug/24
Commented by aleks041103 last updated on 11/Aug/24
$${taken}\:{from}\:{mathstackexchange} \\ $$
Commented by mr W last updated on 11/Aug/24
$${thanks}\:{alot}\:{sir}! \\ $$
Commented by Frix last updated on 12/Aug/24
$$\mathrm{Even}\:\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\mathrm{meant}\:\mathrm{to}\:\mathrm{be} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{6}{k}+\mathrm{7}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{6}{k}+\mathrm{11}\right)^{\mathrm{2}} }\right) \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}… \\ $$$$\mathrm{I}\:\mathrm{guess}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:{a}\pi^{\mathrm{2}} +{b}\:\mathrm{with}\:{a},\:{b}\:\in\mathbb{Q} \\ $$
Commented by Frix last updated on 12/Aug/24
$$\approx\underset{{k}=\mathrm{0}} {\overset{\mathrm{10000}} {\sum}}\:\mathrm{looks}\:\mathrm{like}\:\frac{\pi^{\mathrm{2}} }{\mathrm{9}}−\frac{\mathrm{26}}{\mathrm{25}} \\ $$
Commented by hardmath last updated on 12/Aug/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$