Question 763709
Sum of an Infinite Geometric Series when {{{abs(r)<1}}} is
{{{S=a[1]/(1-r)}}}
1) {{{1024+512+256+128}}}+...
{{{a[1]=1024}}},{{{a[2]=512}}},  {{{r=a[2]/a[1]=512/1024=1/2}}}, since {{{abs(r)<1}}}
{{{S=1024/(1-1/2)=2048}}}
Answer: 2048
2) {{{1/9+1/3+1+3}}}+...
{{{a[1]=1/9}}},{{{a[2]=1/3}}},  {{{r=a[2]/a[1]=(1/3)/(1/9)=3}}}, since {{{r>1}}} 
it is not possible to find the sum 
Answer: No sum
3){{{16+4+1+1/4}}}+...
{{{a[1]=16}}},{{{a[2]=4}}},  {{{r=a[2]/a[1]=4/16=1/4}}}, since {{{abs(r)<1}}}
{{{S=16/(1-1/4)=64/3}}}
Answer: {{{64/3}}}