Question 236799
I'll do the first one to get you started.


# 1



Take note that *[Tex \LARGE 1+3\left(\frac{1}{4}\right)+9\left(\frac{1}{4}\right)^2+27\left(\frac{1}{4}\right)^3+\cdots=1+\left(\frac{3}{4}\right)^1+\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\cdots]



which is a geometric series generated by the sequence *[Tex \LARGE a_{n}=a r^{n-1}] for *[Tex \LARGE n \ge 1]. In this case, {{{a=1}}} and {{{r=3/4}}}


Now recall that an infinite geometric series only converges if {{{abs(r)<1}}}. Since {{{abs(r)=abs(3/4)=3/4=0.75<1}}} holds, this means that this infinite geometric series converges.


In other words, *[Tex \LARGE 1+3\left(\frac{1}{4}\right)+9\left(\frac{1}{4}\right)^2+27\left(\frac{1}{4}\right)^3+\cdots] adds up to some finite number.



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# 2



Take note that *[Tex \LARGE 1+3\left(\frac{1}{5}\right)+9\left(\frac{1}{5}\right)^2+27\left(\frac{1}{5}\right)^3+\cdots=1+\left(\frac{3}{5}\right)^1+\left(\frac{3}{5}\right)^2+\left(\frac{3}{5}\right)^3+\cdots]



which is a geometric series generated by the sequence *[Tex \LARGE a_{n}=a r^{n-1}] for *[Tex \LARGE n \ge 1]. In this case, {{{a=1}}} and {{{r=3/5}}}


Now recall that an infinite geometric series only converges if {{{abs(r)<1}}}. Since {{{abs(r)=abs(3/5)=3/5=0.6<1}}} holds, this means that this infinite geometric series converges.


In other words, *[Tex \LARGE 1+3\left(\frac{1}{5}\right)+9\left(\frac{1}{5}\right)^2+27\left(\frac{1}{5}\right)^3+\cdots] adds up to some constant.





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# 3



Take note that *[Tex \LARGE 1+3\left(\frac{1}{2}\right)+9\left(\frac{1}{2}\right)^2+27\left(\frac{1}{2}\right)^3+\cdots=1+\left(\frac{3}{2}\right)^1+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\cdots]



which is a geometric series generated by the sequence *[Tex \LARGE a_{n}=a r^{n-1}] for *[Tex \LARGE n \ge 1]. In this case, {{{a=1}}} and {{{r=3/2}}}


Now recall that an infinite geometric series only converges if {{{abs(r)<1}}}. Since {{{abs(r)=abs(3/2)=3/2=1.5<1}}} is NOT true, this means that this infinite geometric does NOT converge. So the series diverges. In other words, the infinite series does not add up to some constant number.