(1)
初項1,公比-1/3の等比数列なので
lim[n→∞]{(1-(-1/3)^n)/(1-(-1/3))}
=(1-0)/(1+1/3)
=1/(4/3)
=3/4
に収束
(2)
初項1,公比1/5の等比数列なので
lim[n→∞]{(1-(1/5)^n)/(1-1/5)}
=(1-0)/(4/5)
=1*5/4)
=5/4
に収束
(3)
n項は(-7)^nであり
lim[n→∞](-7)^n=∞となり発散
(4)
Σ[1→∞]{1/((2n-1)(2n+1))}
=Σ[1→∞]{1/2*(1/(2n-1)-1/(2n+1))}
=lim[n→∞]{1/2*(1/1-1/3)+1/2*(1/3-1/5)+…+1/(2n-1)-1/(2n+1)}
=lim[n→∞]{1/2*(1/1-1/3)+1/2*(1/3-1/5)+…+1/2*(1/(2n-1)-1/(2n+1))}
=lim[n→∞]{1/2*(1/1-1/3+1/3-1/5+…+1/(2n-1)-1/(2n+1))}
=lim[n→∞]{1/2*(1-1/(2n+1))}
=1/2*(1-0)
=1/2
