Question 196526
Well first off, if a series diverges, it does not have a sum (ie the if you add up the infinite terms, you won't get a fixed number).



In this problem, however, the series does converge. Here's why:



The sequence of terms: 4, 2, 1, ... can be modeled by the formula:


{{{a[n]=4*(1/2)^n}}} where "n" starts at {{{n=0}}} (note: plug in some values to test this)



Since this equation fits the form {{{a[n]=a*r^n}}} and {{{r=1/2<1}}}, this means that the absolute value of "r" is less than 1. So the series converges.



To find the sum of the infinite series, simply use the formula {{{S=a/(1-r)}}} where "a" is the first term and "r" is given as {{{1/2}}}


So the infinite sum is...


{{{S=4/(1-1/2)=4/(1/2)=4(2)=8}}}



In other words, 


4+2+1+...=8



I hope this makes sense.