Question 83773
In order to find the terms of the series {{{sum( -4(1/3)^(n-1), n=1, infinity )}}} we need to find the terms of the sequence {{{-4(1/3)^(n-1)}}} first.


If we start at n=1, our first number of the sequence is 

{{{-4(1/3)^(1-1)=-4(1/3)^0=-4}}}

If we let n=2, our second number of the sequence is 

{{{-4(1/3)^(2-1)=-4(1/3)^1=-4/3}}}

If we let n=3, our third number of the sequence is 

{{{-4(1/3)^(3-1)=-4(1/3)^2=-4/9}}}

If we let n=4, our fourth number of the sequence is 

{{{-4(1/3)^(4-1)=-4(1/3)^3=-4/27}}}

If we let n=5, our fifth number of the sequence is 

{{{-4(1/3)^(5-1)=-4(1/3)^4=-4/729}}}

If we let n=6, our sixth number of the sequence is 

{{{-4(1/3)^(6-1)=-4(1/3)^5=-4/59049}}}

and so on...




a. write the first four terms of the series

Now that we've generated the sequence of numbers, lets generate the series


So now lets add up the individual terms of the sequence we found earlier

Sum of the first 2 terms of the sequence is 
 {{{-4 / 1+-4 / 3=-16 / 3=-5.33333333333333}}} 
Sum of the first 3 terms of the sequence is
 {{{-4 / 1+-4 / 3+-4 / 9=-52 / 9=-5.77777777777778}}} 
Sum of the first 4 terms of the sequence is
 {{{-4 / 1+-4 / 3+-4 / 9+-4 / 27=-160 / 27=-5.92592592592593}}} 
Sum of the first 5 terms of the sequence is
 {{{-4 / 1+-4 / 3+-4 / 9+-4 / 27+-4 / 81=-484 / 81=-5.97530864197531}}}



So the first four terms are


{{{-16/3}}},{{{-52/9}}},{{{-160/27}}}, {{{-484/81}}}


b. does the series diverge or converge?

If we let the series go on long enough, the series will converge. It's hard to see with the fractions, but the decimal values are clear enough. The series goes 

-5.3333333333333 , -5.77777777777778 ,  -5.92592592592593 , -5.97530864197531...

where the 11th term is -5.99989838947315 and the 32th term is -5.99999999999999. So the series converges to -6. 


c. If the series has a sum, find the sum


To find the sum of an infinite series, use this formula


{{{S=a/(1-r)}}} where S is the sum, a is the first term (in this case -4) and r is the ratio (in this case {{{1/3}}})


{{{S=-4/(1-1/3)}}} plug in a=-4 and {{{r=1/3}}}


{{{S=-4/(3/3-1/3)}}} Make 1 into an equivalent fraction with a denominator of 3


{{{S=-4/(2/3)}}} Combine the fractions in the denominator


{{{S=(-4/1)*(3/2)}}} Multiply the fractions by multiplying the first fraction by the reciprocal of the second


{{{S=-12/2}}} Multiply


{{{S=-6}}} Reduce


So the infinite sum converges to -6