Find the values of x for which the series converges, find the sum of those values.

Hi,

Here's the question,

"Find the values of x for which the series converges. Find the sum of the series for those values of x.

summation(n=0,inf) [(x+3)^n]/2^n"

I wasn't really sure how to approach it. I realized that it can become a convergent geometric series if |x+3| < 2, because that would make the absolute value of the r value less than 1.

Taking this into account, I am proposing that -5< x < -1

I am pretty sure am I right up to this point after looking around online, but I am confused about the second part. I thought maybe you could find the infinite sum when x is -5 and when x is -1 and then subtract the first value from the second value. Is this the correct way to go about it? I got 8/21 as an answer.

Thanks

Re: Find the values of x for which the series converges, find the sum of those values

Quote:

Originally Posted by

**Coop** Hi,

Here's the question,

"Find the values of x for which the series converges. Find the sum of the series for those values of x.

summation(n=0,inf) [(x+3)^n]/2^n"

So

Quote:

I wasn't really sure how to approach it. I realized that it can become a convergent geometric series of |x+3| < 2, because that would make the absolute value of the r value = 1.

Yes, that is a geometric sequence with "common ratio" which must be between -1 and 1 in order that it converge absolutely.

Quote:

Taking this into account, I am proposing that -5< x < -1

Okay, so that -2< x+ 3< 2, -5< x< -1.

Quote:

I am pretty sure am I right up to this point after looking around online, but I am confused about the second part. I thought maybe you could find the infinite sum when x is -5 and when x is -1 and then subtract the first value from the second value. Is this the correct way to go about it? I got 8/21 as an answer.

Thanks

I am afraid you have completely misunderstood this part of the question. It is asking for the sum **as a function of x**, not a difference between two values. You have correctly identified it as a geometric sequence, , with common ratio and first term a= 1. Now, recall that the sum of such a geometric series is .

Re: Find the values of x for which the series converges, find the sum of those values

Quote:

Originally Posted by

**HallsofIvy** I am afraid you have completely misunderstood this part of the question. It is asking for the sum

**as a function of x**, not a difference between two values. You have correctly identified it as a geometric sequence,

, with common ratio

and first term a= 1. Now, recall that the sum of such a geometric series is

.

Isn't the "a" value whatever you have when n=1, so wouldn't "a" be (x+3)/2 and since the whole sequence is in parenthesis, that would be the "r" value as well?

Re: Find the values of x for which the series converges, find the sum of those values

No, the a value is the starting value. Your first term has n = 0, not n = 1.

Re: Find the values of x for which the series converges, find the sum of those values

Quote:

Originally Posted by

**Prove It** No, the a value is the starting value. Your first term has n = 0, not n = 1.

Oh of course, I forgot about the 0, thanks :)

Re: Find the values of x for which the series converges, find the sum of those values

Quote:

Originally Posted by

**Prove It** No, the a value is the starting value. Your first term has n = 0, not n = 1.

So the sum = 1/(1-((x+3)/2)), that's the final answer?

Re: Find the values of x for which the series converges, find the sum of those values

Well, do the **algebra** now! What does that reduce to?

Re: Find the values of x for which the series converges, find the sum of those values

Quote:

Originally Posted by

**HallsofIvy** Well, do the **algebra** now! What does that reduce to?

2/(x-1), right?

Thanks for the help :)

Re: Find the values of x for which the series converges, find the sum of those values

Re: Find the values of x for which the series converges, find the sum of those values

Quote:

Originally Posted by

**Prove It** Try again...

Yeah I forgot about the parenthesis :/ Anyway, thanks.