SOLUTION: Hello! I’m really stumped on this. The question says to consider the infinite geometric series: sigma infinity n=1 -4(1/3)^n-1 and it says the lower limit of the summation

Algebra ->  Sequences-and-series -> SOLUTION: Hello! I’m really stumped on this. The question says to consider the infinite geometric series: sigma infinity n=1 -4(1/3)^n-1 and it says the lower limit of the summation       Log On


   

Question 1141362: Hello! I’m really stumped on this. The question says to consider the infinite geometric series:
sigma infinity n=1 -4(1/3)^n-1
and it says the lower limit of the summation notation is “n=1”. It then says to write the first four terms of the series, asks if the series diverge or converge, and if the series has a sum, find it. I think the series converge, and if I solved for the sum correctly, I got -5.9..., but I can’t figure out the first four terms right. Can you help? And have I solved the rest correctly?
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
Assuming this is right:
+sum%28-4%281%2F3%29%5E%28n-1%29%2C+n=1%2Cinfinity%29+


then your answer for the sum of the first four terms is correct (more precisely, the answer is -160/27 which is about -5.925925...).

The sum of n terms can be found from this:
+S%5Bn%5D+=+a%5B1%5D%2A%281-r%5En%29%2F%281-r%29+
where r is the common ratio (1/3) and a%5B1%5D is the first term (-4) ( so you do need to plug in n=1 to find a%5B1%5D )
Example:


You could also plug in n=1,2,3,4 into the summation and write it all out long-hand but its a lot of extra work.



As n-->infinity, the r%5En term goes to zero, leaving just a%5B1%5D%2F%281-r%29+ as the sum of the infinite progression: